Question

A boy stands on a diving board and tosses a stone into a swimming pool. The stone is thrown from a height of 2.50 m above the water surface with a velocity of 4.00 m/s at an angle of 60.0° above the horizontal. As the stone strikes the water surface, it immediately slows down to exactly half the speed it had when it struck the water and maintains that speed while in the water. After the stone enters the water, it moves in a straight line in the direction of the velocity it had when it struck the water. If the pool is 7.31 m deep, how much time elapses between when the stone is thrown and when it strikes the bottom of the pool?
_____ s

Answer 1

Vyi = 4sin(60) = 3.46m/s
Vxi = 4cos(60) = 2m/s

To find the time the stone is in the air use equation,

df - di = Vyi*t - 0.5*g*t^2
0 = 2.5 + 3.46t -0.5*9.8t^2     =>   t = 1.15s

Now need to find the velocity when the stone hits the water. We know the x-component already (2m/s). So find the y-component.

Vyf = Vyi - g*t = 3.46 - 9.8(1.15) = -7.81m/s
Vxf=Vxi= = 2m/s

The problem says the velocity becomes half, hence

Vyf = -3.9m/s
Vxf = 1m/s

We only care about the vertical component, because the problem asks how long to hit the bottom of the pool, Since velocity is now constant, we can use,

d = vt or t = d/v

t = 7.31/3.9 = 2.03

Total time = 1.15s+2.03 = 3.18s