(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
<u>The given parameters include:</u>
- constant velocity of the elevator, u₁ = 10 m/s
- initial velocity of the ball, u₂ = 20 m/s
- height of the boy above the elevator floor, h₁ = 2 m
- height of the elevator above the ground, h₂ = 28 m
To calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.

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I just did this last semester... I know this is wave height!!
Answer:
a
Explanation:
because it has more energy
Answer:
2.605m
Explanation:
Using the formula for calculating Range (distance travelled in horizontal direction)
Range R = U√2H/g
U is the speed = 4.8m/s
H is the maximum height = ?
g is the acc due to gravity = 9.8m/s²
R = 3.5m
Substitute into the formula and get H
3.5 = 4.8√2H/9.8
3.5/4.8 = √2H/9.8
0.7292 = √2H/9.8
square both sides
0.7292² = 2H/9.8
2H = 0.7292² * 9.8
2H = 5.21
H = 5.21/2
H = 2.605m
Hence the height of the ball from the ground is 2.605m