A boy stands on a diving board and tosses a stone into a swimming pool. The ston
ID: 1333496 • Letter: A
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 7.20 m deep, how much time elapses between when the stone is thrown and when it strikes the bottom of the pool?
Explanation / Answer
We first consider the vertical motion of the stone as it falls toward the water. The initial y velocity component of the stone is
Vyi = vi*sintheta = -4 * sin60 = -3.46 m/s
y co-ordinate is
yf = yi + Vyi*t -1/2*g*t^2
yf = 2.50 -3.46 -4.9t^2
water surface y = 0
4.9 t^2 + 3.46 t -2.50 = 0
t = 0.443 s
Vyf = Vyi -gt
Vyf = -3.46 -9.8*0.443 = -7.81 m/s
After the stone enters to water, its speed, and therefore the magnitude of each velocity component, is reduced by one-half. Thus, the y component of the velocity of the stone in the water is
Vyi = -7.81/2 = -3.91 m/s
this component remains constant until the stone reaches the bottom. As the stone moves through the water, its y coordinate is
yf = yi + Vyt + 1/2*ayt^2
yf = -3.91t
The stone reaches the bottom of the pool when yf = -7.20 m
-7.20 = -3.91t
t = 1.84 s
total time = 1.84 + 0.443
total time = 2.283 s
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