35. A rock is thrown so that it is in the air 4.3 seconds and lands 36 m from its starting point. What was the initial vertical and horizontal speed?

H) sx = vxt vx = sx/t vx = 36 m / 4.3 sec vx = 8.37 m/sec V) sy = (1/2)ayt2 + vyt 0 = 1/2 (-9.8) (4.3)2 + vy vy = 21.1 m/sec

36. With his trusty slingshot, Morris Cumming fires a pebble which is in the air for 5.3 seconds and lands 73 m from where it took off. What were the initial vertical and horizontal components of velocity?

H) sx = vxt vx = sx/t vx = 73 m / 5.3 sec vx = 13.8 m/sec V) sy = (1/2)ayt2 + vyt 0 = 1/2 (-9.8) (5.3)2 + vy vy = 26.0 m/sec

Look over what we are doing in these solutions. First we split each problem into two separate parts: a vertical one and a horizontal one. In the case of projectile motion, the vertical velocity is the thing that gets the ball into the air. The greater the vertical velocity, the longer the ball will stay aloft.

The horizontal velocity has no effect whatsoever on how long the ball stays up, but it's the thing that move the ball forward. We see, then, that the range of a projectile is a combination of the horizontal speed at which the projectile moves toward its target, and the vertical speed which determines how long the projectile is allowed to keep moving.

37. Write the horizontal and vertical equations for each diagram, but do not solve.

a)	b)	c)
d)	e)	f)
a) H) s = vt R = 62 cos40 t V) s = (1/2)at2 + v0t 83 = 1/2(-9.8)t2 + 62 sin40 t	b) H) s = vt R = 55 cos50 t V) s = (1/2)at2 + v0t -90 = 1/2(-32)t2 + 55 sin50 t	c) H) s = vt 29 = v cos62 t V) s = (1/2)at2 + v0t 18 = 1/2(-9.8)t2 + v sin62 t
d) H) s = vt 77 = v cos36 t V) s = (1/2)at2 + v0t 0 = 1/2(-9.8)t2 + v sin36 t	e) H) s = vt 88 = vt V) s = (1/2)at2 + v0t 140 = 1/2(32)t2 + 0	f) H) s = vt 30 = 21 cosØ t V) s = (1/2)at2 + v0t 111 = 1/2(9.8)t2 + 21 sinØ t

38. A B-B takes off at 52° above the horizontal with an initial speed of 36 m/sec. What is its range?

V) sy = (1/2)ayt2 + vyt 0 = 1/2 (-9.8) t2 + 36 sin52 t t = 5.79 sec H) sx = vxt sx = (36 cos52)(5.79) sx = 128 m

39. Grant Wishes is shot out of a cannon at 42° with an initial speed of 11.6 m/sec. How far away must he place the safety net?

V) sy = (1/2)ayt2 + vyt 0 = 1/2 (-9.8) t2 + 11.6 sin42 t t = 1.58 sec H) sx = vxt sx = (11.6 cos42) (1.58) sx = 13.6 m

40. A college freshman, frustrated with finals, releases his tensions by bombarding the adjacent dorm with water balloons. He fires one at 38° from the horizontal with a speed of 21.4 m/sec. How far up the building does it get? The dorm is 13.2 m away.

H) sx = vxt t = s/v 13.2 / 21.4 cos38 = .783 sec V) sy = (1/2)ayt2 + vyt sy = 1/2 (-9.8) (.783)2 + 21.4 sin38 (.783) sy = 7.31 m

41. Since the great troglodyte war of previous pages, colony B has rebuilt and has developed its own technology. It can fire projectiles at different angles, though they all are released at v₀ = 26 m/sec. How far from the top of the cliff will a bomb aimed 22° below the horizontal land?

H) sx = vxt t = sc/vx t= 46 m / 26 cos22 m/sec t = 1.74 sec V) sy = (1/2)ayt2 + vyt sy = 1/2 (-9.8) (1.74)2 + 26 sin22 (1.74) sy = 31.8 m ay + vy are down and so is sy

42**. Where should troglodytes B aim to strike fear into those of colony A who live 88 meters below the surface? Remember, the projectile is fired at 26 m/sec.

V) sy = (1/2)ayt2 + vyt 88 = 1/2 (-9.8) t2 + 26 sinØt H) sx = vxt 88 = 1/2 (9.8) (42 / 26cosØ)2 + 26 cosØ (42 / 26cosØ) The solution requires a programmable calculator: Ø = 52.2°

You can find more general solutions to projectile problems by assigning variables rather than numerical values to the significant quantities.

43. A rock is kicked with velocity v off a cliff of height h. The acceleration of gravity is g.
a) How long will it take to land?
b) How far from the base of the cliff will it land?

a) sy = (1/2)ayt2 h = (1/2)gt2 t = (2h/g)1/2 b) sx = vxt sx = v(2h/g)1/2

44. A disgruntled auto worker pushes a small foreign import off a 22 meter high cliff. If the vehicle lands 13 meters from the base, how fast was it pushed initially?

V) s = (1/2)at2 (Initial vertical velocity is zero) 22 = 1/2 (9.8)t2 t = 2.12 sec H) s = vt 13 = v (2.12) v = 6.14 m/sec

45. A creative chef cracks walnuts by catapulting them into a wall a distance s away. If she releases them at a velocity v and angle q,
a) What are the initial horizontal and vertical velocities?
b) How long is the projectile in the air?
c) How high does it strike the wall?

a) vx = v cosØ vy = v sinØ b) sx = vx t = sx / vx s / v cosØ c) sy = (1/2)ayt2 + vyt = 1/2 (-g) (s / v cosØ)2 + v sinØ (s / v cosØ) sy = (-g / 2) (s2 / v2 cos2Ø) + s(sinØ / cosØ)

46. A disgruntled fast food operator hurls an Egg McMuffin at 18 m/sec, 26° above the horizontal, off a 12 meter high cliff.
a) What are its initial horizontal and vertical velocities?
b) How long will it be in the air?
c) How far from the cliff's base will it land?

a) vx = 18 cos26 = 16.2 m/sec vy = 18 sin26 = 7.89 m/sec b) s = (1/2)at2 + v0t -12 = 1/2 (-9.8)t2 + 18 sin26t t = (18 sin26 + ((18 sin26)2 - 4(4.9)(-12)))1/2 / 2 (4.9) t = 2.57 sec c) s = vt = (18 cos26) (2.57) s = 41.5 m

47. A projectile is fired vertically up at 35 m/sec. How long will it take to reach the height of 40 m? Is it meaningful to have two solutions?

V) This is on the way up s = (1/2)at2 + v0t 40 = 1/2(-9.8)t2 + 35t 4.9t2 - 35t + 40 = 0 On the way down t = (35 + ((35)2 - 4(4.9) (40)))1/2 / 2 (4.9) t = 1.43, 5.71 sec

48. A projectile is fired up at 27 m/sec. How long will it be in flight if a bird catches it in mid air, while it is falling, 25 m from the ground?

V) s = (1/2)at2 + v0t 25 = 1/2(-9.8)t2 + 27t 4.9t2 - 27t + 25 = 0 t = (27 + ((27)2 - 4(4.9) (25)))1/2 / 2 (4.9) t = 1.18, 5.71 sec

49. A ball on a cliff is thrown vertically down at 12 ft/sec. If the cliff is 77 feet high, how long will it take to hit the ground?

s = (1/2)at2 + v0t 77 = 1/2(32)t2 + 12t 16t2 + 12t - 77 = 0 t = (-12 + ((12)2 - 4(16) (-77)))1/2 / 2 (16) t = 1.85 sec

50. A ball is thrown up at 19 ft/sec off the same 77 foot high cliff you learned to know and love in problem #49. How long will it take to hit the ground?

s = (1/2)at2 + v0t -77 = 1/2(-32)t2 + 19t t = (19 + ((-19)2 - 4(16) (-77)))1/2 / 2 (16) t = 2.87 sec

51. This time the ball is thrown off the edge of the cliff at 40 ft/sec, 28° above the horizontal. How long will it take to reach the ground? (The cliff is still 77 feet high.)

Only the vertical component matters vy = 40 sin28 s = (1/2)at2 + v0t -77 = 1/2(-32)t2 + 40 sin 28t 16t2 - 40 sin28t - 77 = 0 t = 2.86 sec

52. Now a ball is fired at 37 m/sec, 41° above the horizontal, up onto a cliff that is 26 m high. How far does it travel horizontally?

V) s = (1/2)at2 + v0t 26 = 1/2(-9.8)t2 + 37 sin41t 4.9t2 - 37 sin41t + 26 = 0 t = 1.57, 3.39 sec H) s = vt s = (37 cos41) (3.39) s = 94.6 m

53. Finally, a ball is fired at 67° above the horizontal onto a cliff 38 m high, with an initial velocity of 32 m/sec. How far does it move horizontally?

V) s = (1/2)at2 + v0t 26 = 1/2(-9.8)t2 + 32 sin67t 4.9t2 - 32 sin67t + 38 = 0 t = 1.87, 4.14 sec H) s = vt s = (32 cos67) (4.14) s = 51.7 m

54. A bridge is a height h above the ground.
a) How long will it take a rock to fall if released from rest off the bridge? (gravity is g)
b) How long will it take if thrown down at velocity v₀?
c) How long will it take if thrown up at v₀?

a) s = (1/2)at2 h = (1/2)gt2 t = (2h/g)1/2 b) s = (1/2)at2 + v0t [Since all quantities are down, we can assign them all a positive value.] h = (1/2)gt2 + v0t (g/2)t2 + v0t - h = 0 t = (-v0 + (v02 - 4(g/2) (-h)))1/2 / 2(g/2) t = (-v0 + (v02 + 2gh)))1/2 / g c) Here, v0 is up, g and h are down. -h = 1/2 (-g)t2 + v0t (g/2)t2 - v0t - h = 0 t = (v0 + (v02 + 2gh)))1/2 / g

55. A rock is thrown at velocity v₀ at angle q on level ground.
a) What is its initial vertical velocity?
b) How long will it remain in the air?
c) How far will it get before hitting the ground?

a) vy = v0 sinØ b) s = (1/2)at2 + v0t 0 = 1/2(-g)t2 + v0 sinØt t = 2v0 sinØ / g c) s = vxt (v0 cosØ) 2v0 / g

56. Two cars head in the same direction. Car 1 moves with initial velocity v1 and accelerates at rate a. Car 2 moves with constant velocity v², and has a head start of length L. When will the cars meet?

Position of car 1: s1 = (1/2)at2 + v1t Position of car 2: s2 = vt + L Collision when s1 = s2 (1/2)at2 + v1t = v0t + L (a/2)t2 + (v1 - v2)t - L = 0 t = (-(v1 - v2) + ((v1 - v0)2 - 4(a/2) (-L))1/2 / 2 (a/2)) t = ((v2 -v1) + ((v1 - v2)2 + 2aL)))1/2 / a)

57. A rock is kicked horizontally off a tall cliff at velocity v₀. What will its angle and speed be at time t?

H) vx = v0 For the entire trip. V) vy = -gt v = (v02 + g2t2)1/2 = tan-1(gt / v0)

58. A car accelerates at a constant rate a.
a) How long will it take to go a distance s?
b) How fast will it be going after traveling a distance s?

a) s = (1/2)at2 t = (2s/a)1/2 b) v = at v = a(2s/a)1/2 v = (2as)1/2 OR v2 = 2as

This equation, v² = 2as, is the fourth and final kinematic equation. It, like the others, should be memorized. The four are:

s = vt

s = 1/2 at2

v = at

v2 = 2as

59*. A particle with initial velocity v₀ accelerates at rate a. How fast will it be going after traveling a distance s?

s = (1/2)at2 + v0t (a/2)t2 + v0t - s = 0 t = (-v0 + ((v0)2 - 4(a/2) (-s)) / a)1/2 v = at + v0 v = a(-v0 + ((v0)2 - 2as)1/2 / a) + v0 v = (v02 + 2as)1/2 Square both sides: v2 = 2as + v02

60. Rosa Corn throws a banana vertically upward at 24 m/sec. How fast is it traveling at a height of 19 m above ground?

v2 = 2as + v02 v2 = 2 (-9.8) (19) + 242 v2 = 204 v = 14.3 m/sec

61. A champagne cork pops vertically upward at 29 ft/sec. To what maximum height will it rise? (Think What is its final velocity at the peak of its flight?)

v2 = 2as + v02 At the max height, v = 0 02 = 2 (-32) (s) + 292 s = 13.1 ft

62. At a race track a car is clocked at 15 ft/sec at one point, and at 47 ft/sec 40 feet farther down the track. What is its acceleration?

v2 = 2as + v02 472 = 2 (a) (40) + 152 a = 24.8 ft/sec2

63. A bolt drops out of a bridge and falls 117 m to the river below. How fast is it traveling when it hits the water?

v2 = 2as + v02 v2 = 2 (9.8) (117) + 02 v2 = 2293 v = 47.9 m/sec

64. Arthur More watches the splash from the bolt in problem 63. He tries for a bigger splash by throwing a rock vertically downward at 27 m/sec. How fast will it strike the water?

v2 = 2as + v02 v2 = 2 (9.8) (117) + 272 v2 = 3022 v = 55.0 m/sec note: Arthur would get the same splash if he'd thrown the rock up at 27 m/sec.

65*. A bowling ball is flung at 106 ft/sec, 65° above the horizontal, striking a wall a distance L away. How high up the wall will it hit?

H) s = vt 170 = 106 cos65t t = 3.79 sec V) s = (1/2)at2 + v0t h = 1/2 (-32) (3.79)2 + (106 sin65) (3.79) h = 134 ft

66*. Two walls are separated by a distance L. A projectile fired at velocity v₀, downward at q as shown, lands a distance h below the top. Express q in terms of v₀, L, h, and g.

H) sx = vxt L = v0 cosØt t = L / v0 cosØ V) sy = (1/2)ayt2 + vyt h = (-g/2) (L / v0 cosØ)2 + (v0 sinØ) (L / v0 cosØ)

Note that I have assigned positive values to all quantities in the vertical equation. Because gravity, initial velocity and final location are all in the same direction, they must be assigned the same sign -- all negative or all positive.

67**. Some happy college freshmen decide to have a water balloon fight in the hallway of their dorm. The ceiling is of height h, and the balloons are launched at velocity v₀. At what angle will the balloons just graze the ceiling?

Note that when sy = h, vy = 0 (Think about it!) Use our newest kinematic equation: v2 = 2as + v02 02 = 2(-g)h + (v0 sinØ)2 Solving for Ø: Ø = sin-1((2gh) / v0)1/2