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

Explanation / Answer

A stone is thrown straight up at a velocity of 4 * sin(60) = 2*sqrt(3) meters / sec What are you using as acceleration due to gravity? I'll use 9.81 m / sec^2 velocity = -9.81 t + c . . . at t(0) velocity = 2*sqrt(3) so c = 2*sqrt(3) velocity = - 9.81t + 2*sqrt(3) distance = - 4.905 t^2 + 2sqrt(3) t + c . . . at t(0) distance = 2.5 meters (above water) so c = 2.5 distance from water surface = - 4.905 t^2 + 2sqrt(3) t + 2.5 distance = 0 when stone hits the water 0 = - 4.905 t^2 + 2sqrt(3) t + 2.5 t = - 0.443359 (invalid) or t = 1.1496 velocity at t = 1.1496 -9.81 * 1.1496 + 2*sqrt(3) = -7.81347 meters / sec (minus indicates direction) Stone slows to half that and travels 4.74 meters 7.81347 / 2 = 3.906735 meters / sec (4.74 meters) / (3.906735 meters / sec) = 1.213 seconds from hitting the water till reaching the bottom. To this we add the time from thrown to hitting the water 1.1496 + 1.213 = 2.3628 seconds

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.