Projectile Motion:- Projected at an angle up from a height

Q.1. Theo threw a water balloon from the top of a cliff 26 metres high at an angle of 10 degrees up from the horizontal with a speed of 86 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 speed of the balloon just before it hit the ground? (nearest m/s)
l. At what angle to the horizontal did the ball hit the ground below? (nearest degree)

Q.2. Petra threw a water balloon from the top of a cliff 69 metres high at an angle of 60 degrees up from the horizontal with a speed of 87 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 speed of the balloon just before it hit the ground? (nearest m/s)
l. At what angle to the horizontal did the ball hit the ground below? (nearest degree)

Answers

Q.1. Theo threw a water balloon from the top of a cliff 26 metres high at an angle of 10 degrees up from the horizontal with a speed of 86 m/s.
a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
Horizontal component = v cosθ
= 86 cos10= 84.69347 m/s = 84.7m/s (nearest 0.1 m/s)

b. What was the initial vertical component of the velocity? (nearest 0.1m/s)
vertical component = v sinθ
= 86 sin10 = 14.93374m/s = 14.9 m/s (nearest 0.1 m/s)

c. How long did it take the water balloon to reach maximum height?
v = u + at
0 = 14.9 - 9.8t
9.8t = 14.9
t = 14.9/9.8 = 1.52041 = 1.5s (nearest 0.1s)

d. What was the maximum height reached above the ground below the cliff?
v2 = u2 + 2 a s
0 = 14.92 + 2 x (-9.8) x s (Note that the initial velocity and g are in opposite directions and have opposite signs.)
19.6s = 14.92 = 222
s = 222/19.6 = 11.3265m above clifftop = (11.3 + 26)m = 37m above ground below cliff (nearest m)

e. What was the velocity of the balloon at maximum height? (nearest 0.1m/s)
84.7 m/s horizontally: 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
37 = 0 + ½ x 9.8 x t2
37 = 4.9 t2
t = √ (37/4.9) = 2.7479s = 2.7s (nearest 0.1s)

g. What was the total time of flight? (nearest 0.1s)
Total time of flight = time up + time down
= 1.5 + 2.7 = 4.2s

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
= 84.7 x 4.2 = 355.74 = 356m (nearest m)

i. What was the horizontal component of the balloon's velocity just before it hit the ground? (nearest 0.1m/s)
84.7 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 (Speed attained by falling from maximum height with initial velocity zero)
v2 = 0 + 2 x 9.8 x 37 = 725.2
v = √725.2 = 26.9295m/s = 26.9 m/s (to nearest 0.1m/s)

k. What was the speed of the balloon just before it hit the ground? (nearest 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.
Final horizontal component: 84.7 m/s Final vertical component: 26.9m/s speed2 = 84.72 + 26.92
speed2 = 7174.1 + 723.7= 723.6
speed = √(7897.7) =88.869 = 89m/s nearest m/s

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

Direction: tanθ = vertical component of velocity/horizontal component of velocity
Final horizontal component: 84.7 m/s Final vertical component: 26.9m/s tanθ = 26.9/84.7 = 0.31759 θ = tan-10.31759 = 17.619o= 18o (nearest deg)


Q.2. Petra threw a water balloon from the top of a cliff 69 metres high at an angle of 60 degrees up from the horizontal with a speed of 87 m/s.
a. What was the initial horizontal component of the velocity? (nearest 0.1m/s)
Horizontal component = v cosθ
= 87 cos60= 43.5 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
= 87 sin60 = 75.3442m/s = 75.3 m/s (nearest 0.1 m/s)

c. How long did it take the water balloon to reach maximum height?
v = u + at
0 = 75.3 - 9.8t
9.8t = 75.3
t = 14.9/9.8 = 7.68367 = 7.7s (nearest 0.1s)

d. What was the maximum height reached above the ground below the cliff?
v2 = u2 + 2 a s
0 = 75.3sup>2 + 2 x (-9.8) x s (Note that the initial velocity and g are in opposite directions and have opposite signs.)
19.6s = 75.32 = 5670
s = 5670/19.6 = 289.29m above clifftop = (289.29 + 69)m = 358m above ground below cliff (nearest m)

e. What was the velocity of the balloon at maximum height? (nearest 0.1m/s)
43.5 m/s horizontally: 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
358 = 0 + ½ x 9.8 x t2
358 = 4.9 t2
t = √ (358/4.9) = 73.06122s = 73.1s (nearest 0.1s)

g. What was the total time of flight? (nearest 0.1s)
Total time of flight = time up + time down
= 7.7 + 73.1 = 80.8s

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
= 43.5 x 80.8 = 3514.8 = 3515m (nearest m)

i. What was the horizontal component of the balloon's velocity just before it hit the ground? (nearest 0.1m/s)
43.5 m/s horizontally: same as answers (a) and (e) 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 (Speed attained by falling from maximum height with initial velocity zero)
v2 = 0 + 2 x 9.8 x 358 = 7016.8
v = √7016.8 = 283.76634m/s = 283.8 m/s (to nearest 0.1m/s)

k. What was the speed of the balloon just before it hit the ground? (nearest 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.
Final horizontal component: 43.5 m/s Final vertical component: 283.8m/s speed2 = 43.52 + 283.82
speed2 = 1892.25 + 80542= 82434
speed = √(82434) = 287.1132 = 287 m/s nearest m/s

l. At what angle to the horizontal did the ball hit the ground below? (nearest degree)
Direction: tanθ = vertical component of velocity/horizontal component of velocity
Final horizontal component: 43.5 m/s Final vertical component: 283.8 m/s tanθ = 283.8/43.5 = 6.5241 θ = tan-16.5241 = 81.2857o= 81o (nearest deg)