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

Answer 1

Abbly the equation of motion. s = ut + 1 g t ² U = Usintoo g. - 9.0 m/s. so (usinbolt) + .(-9-0) (+)? 2.5 = (u)sinco (H) - 1 19.8) ? 2.5 = 3.46464) - 64.97€) =) 4.972-3.464 € +2: 5 0 solve for t, we get to 1.154 Let t' is the time reach ground back, Uy= utat = (usin6o) + at LUX : (3 - 4 ) +- 4) 1.). --3-7.00 ml U U los o = u cos 60 = 4 x 1 = 2m/s. Cal veloh, V. Tuy + ) -ouỷ Kyº u= 8.06 mis __

Find y = 0.06 = 4:03 m/s Depth of the pool, Z n 2.69m. time t = Depth = 2.6 = 0.667 -- 1/2__ 4.03 hotel Hime (t) = 1.15 + 0.667 = 1.017. a.

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