## Projectile Motion:- Projected at an angle down from a height.

Q.1. Joan threw a water balloon from the top of a cliff 35 metres high at an angle of 20 degrees down from the horizontal with a speed of 64 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 balloon's velocity just before it hit the ground? (nearest 0.1m/s)
d. What was the vertical component of the balloon's velocity just before it hit the ground? (nearest 0.1m/s)
e. How long did it take the balloon to reach the ground below? (nearest 0.1s)
f. How far from the base of the cliff did the balloon first hit the ground? (nearest m)
g. What was the speed of the balloon just before it hit the ground? (nearest m/s)
h. At what angle to the horizontal did the water balloon hit the ground below? (nearest degree)

Q.2. Pam threw a discus from the top of a cliff 71 metres high at an angle of 10 degrees down from the horizontal with a speed of 85 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. Lyn threw a water ball from the top of a cliff 49 metres high at an angle of 50 degrees down from the horizontal with a speed of 50 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 ball's velocity just before it hit the ground? (nearest 0.1m/s)
d. What was the vertical component of the ball's velocity just before it hit the ground? (nearest 0.1m/s)
e. How long did it take the ball to reach the ground below? (nearest 0.1s)
f. How far from the base of the cliff did the ball first hit the ground? (nearest m)
g. What was the speed of the ball just before it hit the ground? (nearest m/s)
h. At what angle to the horizontal did the ball hit the ground below? (nearest degree)

Q.1. Joan threw a water balloon from the top of a cliff 35 metres high at an angle of 20 degrees down from the horizontal with a speed of 64 m/s.
a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
vx = v cosθ
vx = 64 cos 20 = 60.14 m/s = 60.1 m/s (nearest m/s)

b. What was the initial vertical component of the velocity? (nearest 0.1m/s)
vy = v sinθ
vy = 64 sin 20 = 21.89 m/s = 21.9 m/s (nearest m/s)

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

d. 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 = 21.92 + 2 x 9.8 x 35
v2 = 479.6 + 686
v2 = 1165.6
v = √(1165.6) = 34.14m/s

e. How long did it take the balloon to reach the ground below? (nearest 0.1s)
v = u + at
34.14 = 21.9 + 9.8t
34.14 - 21.9 = 9.8t
t = (34.14 - 21.9)/9.8
t = 1.24898 = 1.2 s (nearest 0.1s)

f. How far from the base of the cliff did the balloon first hit the ground? (nearest m)
Range = Horizontal velocity x time of flight
Range = 60.1 x 1.2 = 72.12 = 72m (nearest m)
(Note that taking horizontal velocity and time to 2 decimal places gives 60.14 x 1.25 = 75m)

g. What was the speed of the balloon just before it hit the ground? (nearest m/s)
Horizontal component of velocity = 60.1 m/s
Vertical component of velocity = 34.14 m/s
These two components are at right angles to each other so the magnitude of the vector sum is found by Pythagoras' theorem.
speed2 = 60.12 + 34.142
speed2 = 3612 + 1165 = 4777
speed = √(4777) = 69.1 = 69m/s nearest m/s

h. At what angle to the horizontal did the water balloon hit the ground below? (nearest degree)
tanθ = vertical component of velocity/horizontal component of velocity
tanθ = 34.14/60.1 = 0.56805
θ = 29.6deg = 30 deg nearest degree

* * * * * Q.2. Pam threw a discus from the top of a cliff 71 metres high at an angle of 10 degrees down from the horizontal with a speed of 85 m/s.
a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
vx = v cosθ
vx = 85 cos 10 = 83.7086 m/s = 83.7 m/s (nearest m/s)

b. What was the initial vertical component of the velocity? (nearest 0.1m/s)
vy = v sinθ
vy = 85 sin 10 = 14.7601 m/s = 14.8 m/s (nearest m/s)

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

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 = 14.82 + 2 x 9.8 x 71
v2 = 219.04 + 1391.6
v2 = 1610.64
v = √(1610.64) = 40.13278m/s = 40.1m/s (1 d.p.)

e. How long did it take the discus to reach the ground below? (nearest 0.1s)" + "
v = u + at
40.1 = 14.8 + 9.8t
40.1 - 14.8 = 9.8t
t = (40.1 - 14.8)/9.8
t = 2.5816 = 2.6 s (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 = 83.7 x 2.6 = 217.62 = 72m (nearest m)
g. What was the speed of the discus just before it hit the ground? (nearest m/s)
Horizontal component of velocity = 83.7 m/s
Vertical component of velocity = 14.8 m/s
These two components are at right angles to each other so the magnitude of the vector sum is found by Pythagoras' theorem.
speed2 = 83.72 + 14.82
speed2 = 7006 + 219 = 7225
speed = √(7225) = 85m/s nearest m/s

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

tanθ = vertical component of velocity/horizontal component of velocity
tanθ = 14.8/83.7 = 0.1768
θ = 10.0deg = 10 deg nearest degree

Q.3. Lyn threw a water ball from the top of a cliff 49 metres high at an angle of 50 degrees down from the horizontal with a speed of 50 m/s.

a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
vx = v cosθ
vx = 50 cos 50 = 32.13938 m/s = 32.1 m/s (nearest 0.1m/s)

b. What was the initial vertical component of the velocity? (nearest 0.1m/s)
vy = v sinθ
vy = 50 sin 50 = 38.3022 m/s = 30.3 m/s (nearest 0.1m/s)

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

d. What was the vertical component of the ball's velocity just before it hit the ground? (nearest 0.1m/s)
v2 = u2 + 2 a s
v2 = 30.32 + 2 x 9.8 x 49
v2 = 918.09 + 960.4
v2 = 1878.49
v = √(1878.49) = 43.34m/s
v = 43.3m/s (nearest 0.1m/s
e. How long did it take the ball to reach the ground below? (nearest 0.1s)
v = u + at
43.3 = 30.3 + 9.8t
43.3 - 30.3 = 9.8t
t = (43.3 - 30.3)/9.8
t = 1.3265 = 1.3 s (nearest 0.1s)

f. How far from the base of the cliff did the balloon first hit the ground? (nearest m)
Range = Horizontal velocity x time of flight
Range = 32.1 x 1.3 = 41.72 = 42m (nearest m)
g. What was the speed of the ball just before it hit the ground? (nearest m/s)
Horizontal component of velocity = 32.1 m/s
Vertical component of velocity = 30.3 m/s
These two components are at right angles to each other so the magnitude of the vector sum is found by Pythagoras' theorem.
speed2 = 32.12 + 30.32
speed2 = 1030.4 + 918.1 = 1948.5
speed = √(1948.5) = 44.14 = 44m/s nearest m/s

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

tanθ = vertical component of velocity/horizontal component of velocity
tanθ = 30.3/32.1 = 0.9439
θ = 43.3476deg = 43 deg nearest degree