## Projectile Motion:- Variety of questions from a height

This is an example worksheet with worked solutions to be used with the regenerating worksheets "CombinedFromHeight".

Q.1. Bruno fired a gun horizontally from the top of a cliff 39 metres high. The bullet left the barrel of the gun with a speed of 79 m/s.
a. How long did it take for the bullet to hit the ground? (nearest 0.1s)
b. How far from the base of the cliff did the bullet hit the ground? (nearest m)

Q.2. Ruby threw a discus from the top of a cliff 68 metres high at an angle of 40 degrees down from the horizontal with a speed of 74 m/s.
a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
b. What was the initial vertical component of the velocity? (nearest 0.1m/s)
c. What was the horizontal component of the discus' velocity just before it hit the ground? (nearest 0.1m/s)
d. What was the vertical component of the discus' velocity just before it hit the ground? (nearest 0.1m/s)
e. How long did it take the discus to reach the ground below? (nearest 0.1s)
f. How far from the base of the cliff did the discus first hit the ground? (nearest m)
g. What was the speed of the discus just before it hit the ground? (nearest m/s)
h. At what angle to the horizontal did the discus hit the ground below? (nearest degree)

Q.3. John threw a water balloon from the top of a cliff 72 metres high at an angle of 60 degrees up from the horizontal with a speed of 62 m/s.
a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
b. What was the initial vertical component of the velocity? (nearest 0.1m/s)
c. How long did it take the water balloon to reach maximum height?
d. What was the maximum height reached above the ground below the cliff?
e. What was the velocity of the balloon at maximum height? (nearest 0.1m/s)
f. How long did it take the balloon to fall from maximum height to the ground below? (nearest 0.1s)
g. What was the total time of flight? (nearest 0.1s)
h. How far from the base of the cliff did the balloon first hit the ground? (nearest m)
i. What was the horizontal component of the balloon's velocity just before it hit the ground? (nearest 0.1m/s)
j. What was the vertical component of the balloon's velocity just before it hit the ground? (nearest 0.1m/s)
k. What was the velocity of the balloon just before it hit the ground? (nearest m/s)

Q.1. Bruno fired a gun horizontally from the top of a cliff 39 metres high. The bullet left the barrel of the gun with a speed of 79 m/s.
a. How long did it take for the bullet to hit the ground? (nearest 0.1s)

Note that the initial vertical component of the velocity is zero.
s = ut + ½ at2
u is zero so the equation becomes s = ½ at2
39 = ½ x 9.8 x t2
t2 = 39/4.9 = 7.95918
t = √(7.95918) = 2.821 = 2.8 s (nearest 0.1s)

b. How far from the base of the cliff did the bullet hit the ground? (nearest m)

Range = horizontal velocity x time of flight
range = 79 x 2.8 = 221.2 = 221m (nearest m)

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Q.2. Ruby threw a discus from the top of a cliff 68 metres high at an angle of 40 degrees down from the horizontal with a speed of 74 m/s.
a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
Horizontal component = v cosθ
= 74 cos40 = 56.687m/s = 56.7 m/s (nearest 0.1 m/s)

b. What was the initial vertical component of the velocity? (nearest 0.1m/s)
vertical component = v sinθ
= 74 sin40 = 47.566m/s = 47.6 m/s (nearest 0.1 m/s)

c. What was the horizontal component of the discus' velocity just before it hit the ground? (nearest 0.1m/s)
Horizontal component is unchanged since there are no horizontal forces acting.
Horizontal component of velocity just before hitting ground = 56.7m/s

d. What was the vertical component of the discus' velocity just before it hit the ground? (nearest 0.1m/s)
v2 = u2 + 2 a s
v2 = 47.62 + 2 x9.8 x 68 = 3598.56
v = √(3598.56) = 59.988
v = 60.0m/s (nearest 0.1s)

e. How long did it take the discus to reach the ground below? (nearest 0.1s)
v = u + at
60 = 47.6 + 9.8t
t = (60 - 47.6)/9.8 = 1 265 s = 1.3s (nearest 0.1s)

f. How far from the base of the cliff did the discus first hit the ground? (nearest m)
range = horizontal velocity x time of flight
range = 56.7 x 1.3 = 73.71 m = 74m (nearest m)

g. What was the speed of the discus just before it hit the ground? (nearest m/s)
speed = magnitude of final velocity = vector sum of vertical and horizontal velocities
Since vertical and horizontal velocities are at right angles, then Pythagoras' theorem may be applied.
v2 = vx2 + vy2
v2 = 56.72 + 47.62
= 5480.65 v = &radic(5480.65) = 74.03 = 74 m/s (nearest m/s)

h. At what angle to the horizontal did the discus hit the ground below? (nearest degree)

tanθ = (vertical velocity) / (horizontal velocity) = 47.6/56.7
θ = 40.01o = 40o (nearest deg.)

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Q.3. John threw a water balloon from the top of a cliff 72 metres high at an angle of 60 degrees up from the horizontal with a speed of 62 m/s.
a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
Horizontal component = v cosθ
= 62 cos60= 31.0 m/s (nearest 0.1 m/s)

b. What was the initial vertical component of the velocity? (nearest 0.1m/s)
vertical component = v sinθ
= 62 sin60 = 53.69357m/s = 53.7 m/s (nearest 0.1 m/s)

c. How long did it take the water balloon to reach maximum height?
v = u + at
0 = 53.7 - 9.8t
9.8t = 53.7
t = 53.7/9.8 = 5.47959 = 5.5s (nearest 0.1s)

d. What was the maximum height reached above the ground below the cliff?
v2 = u2 + 2 a s
0 = 53.72 + 2 x (-9.8) x s (Note that the initial velocity and g are in opposite directions and have opposite signs.) 19.6s = 53.72 = 2884
s = 2884/19.6 = 147m above clifftop = (147 + 72)m = 219m above ground below cliff

e. What was the velocity of the balloon at maximum height? (nearest 0.1m/s)
31 m/s same as answer (a) since there are no horizontal forces acting. Horizontal component is constant. Vertical component is zero.

f. How long did it take the balloon to fall from maximum height to the ground below? (nearest 0.1s)
s = ut + ½ at2
219 = 0 + ½ x 9.8 x t2
219 = 4.9 t2
t = √ (219/4.9) = 6.6853s 6.7s (nearest 0.1s)

g. What was the total time of flight? (nearest 0.1s)
Total time of flight = time up + time down
= 5.5 + 6.7 = 12.2s

h. How far from the base of the cliff did the balloon first hit the ground? (nearest m)
Range = Horizontal velocity x time of flight
= 31 x 12.2 = 378.2 = 378m (nearest m)

i. What was the horizontal component of the balloon's velocity just before it hit the ground? (nearest 0.1m/s)
31 m/s same as answer (a) since there are no horizontal forces acting. Horizontal component is constant.

j. What was the vertical component of the balloon's velocity just before it hit the ground? (nearest 0.1m/s)
v2 = u2 + 2 a s
v2 = 0 + 2 x 9.8 x 219 = 4292.4
v = √4292.4 = 65.51m/s = 65.5 m/s (to nearest 0.1m/s)

k. What was the velocity of the balloon just before it hit the ground? (nearest m/s)

Velocity is a vector so we need both the speed and the direction.
Horizontal component of velocity = 31m/s
Vertical component of velocity = 65.5 m/s
Speed:These two components are at right angles to each other so the magnitude of the vector sum is found by Pythagoras' theorem.
speed2 = 312 + 65.52
speed2 = 961 + 4290.25= 5251.25
speed = √(5251.25) = 72.4655 = 72m/s nearest m/s

Direction: tanθ = vertical component of velocity/horizontal component of velocity
tanθ = 65.5/31 = 2.1129 θ = tan-12.1129 = 64.67o= 65o (nearest deg)
Velocity = 72m/s at an angle of 65o down from the horizontal.

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