Answer:
A.) 4.81 seconds
B.) 44.6 m/s
Explanation:
He begins his dive by jumping up with a velocity of 5 (m/s).
Let us first calculate the maximum height reached by using third equation of motion
V^2 = U^2 - 2gH
At maximum height, V = 0
0 = 5^5 - 2 × 9.8H
19.6H = 25
H = 25 /19.6
H = 1.28 m
The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.
Where height h = 1.28 + 100 = 101.28 m
h = Ut + 1/2gt^2
As the diver drop from maximum height, U = 0
101.28 = 1/2 × 9.8 × t^2
4.9t^2 = 101.28
t^2 = 101.28/4.9
t^2 = 20.669
t = sqrt ( 20.669)
t = 4.55s
As the diver jumped up, the time taken to reach the maximum height will be
Time = 1.28 / 5 = 0.256
The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds
B.) Velocity right before he hits the water will be
V^2 = U^2 + 2gH
But U = 0
V^2 = 2 × 9.8 × 101.28
V^2 = 1985.09
V = 44.6 m/s
