A pebble is released from rest at a certain height and falls freely, reaching an impact speed of `4 m//s` at the floor. Next, the pebble is thrown down with an initial speed of `3 m//s` from the same height. What is its speed at the floor ?

Options

A. `4 m//s`
B. `5 m//s`
C. `6 m//s`
D. `7 m//s`

Answer

B

Explanation

Take down as the positive direction. Since the pebble is released from rest, `v_f^2 = v_i^2 + 2 a Delta y` becomes
`v_f^2 = (4 m//s)^2 = 0^2 + 2gh`
Next, when the pebble is thrown with speed `3.0 m//s` from the same height `h`, we have
`v_f^2 = (3 m//s)^2 + 2gh = (3 m//s)^2 +(4 m//s)^2`
`rArr v_f = 5 m//s`.