Question

4. A boy stands on a diving board and tosses a stone into a swimming pool. The stone is thrown from a height of 2.8 m above the water surface with a velocity of 5.6 m/s at 72 degrees above the horizontal.

a) First determine the time it is airborne. (I got 1.48 sec)

b) Now find its impact velocity. (Vx=2 m/s and Vy=9.17 m/s)

As the stone impacts the water, it instantly slows down to half its previous velocity, and maintains that velocity (both speed and direction) as it sinks to the bottom of the pool, which is 4.0 m deep.

c) Find the time it takes to reach the bottom. (1.93 sec)

Bonus +3 What is the total horizontal displacement of the stone over its entire trip? I dont know how to do this one

Answer 1

Answer :

First, you can simplify the whole problem by removing the horizontal component.
A stone is thrown straight up at a velocity of 5.6 * sin(72) = 5.33 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 = 5.33 so c = 5.33
velocity = - 9.81t + 5.33

distance = - 4.905 t^2 + 5.33 t + c
. . . at t(0) distance = 2.8 meters (above water) so c = 2.8
distance from water surface = - 4.905 t^2 + 5.33 t + 2.8

distance = 0 when stone hits the water

0 = - 4.905 t^2 + 5.33 t + 2.8
t = -0.3873 (invalid)
or
t = 1.473

velocity at t = 1.473

-9.81 * 1.473 + 5.33 = -9.105 meters / sec

(minus indicates direction)

Stone slows to half that and travels 4.0 meters

9.105 / 2 = 4.5527meters / sec
(4.0 meters) / (4.5527meters / sec) =
0.87859 seconds from hitting the water till reaching the bottom.

To this we add the time from thrown to hitting the water

1.473 + .0.87859 = 2.35 seconds