A boy stands on a diving board and tosses a stone into a swimming pool. The ston

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 3.00 m deep, how much time elapses between when the stone is thrown and when it strikes the bottom of the pool?

Explanation / Answer

Since the stone is a projectile, the horizontal component of its velocity remains constant in the air. The vertical is accelerated by gravity. Use trig: Initial y-velocity = 4sin(60) = 3.46m/s Initial x-velocity = 4cos(60) = 2m/s (constant) To find the time the stone is in the air you need to find the time it takes to go up, stop, come back down, and fall another 2.5m to the water. This is determined from the y-direction motion only: d = do + Vyo*t - 0.5*g*t^2 0 = 2.5 + (3.46)t - 4.9t^2 Use quadratic equation to get 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. Note the stone should be falling faster now, right? Vy = Vo - g*t = 3.46 - 9.8(1.15) = -7.81m/s Vx = 2m/s The problem says the velocity becomes half, so: Vy = -3.9m/s Vx = 1m/s We only care about the vertical component, because the problem asks how long to hit the bottom of the pool, and we don't care how far over (left/right) it goes. Since velocity is now constant, we can use: d = vt or t = d/v So t = (2.39m)/(3.9m/s) = 0.61s

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