Physics Homework

Question

A rock is tossed straight up with a speed of 20 m/s. When it returns, it falls into a hole 10 m deep.

a) What is the rocks velocity as it hits the bottom of the hole?

b) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Answer 1

First let's find the height the rock reaches on the way up. From energy methods,

20 m/s = sqrt(2*g*h) where g = acceleration due to gravity = 9.8 m/s2

So.

400 = 2*9.8*h or h = 20.41 m and since h = (1/2)gt2 the time going up is

t = sqrt(2*h/9.8) = 2.041 s

Now let's add 10 meters to 20.41 = 30.41 to get the drop distance. The the time will be

t =sqrt(2*30.41/9.8) = 2.49 s and the velocity at the bottom of the hole will be

v = sqrt(2*9.8*30.41) = 24.41 m/s