Question

A student throws a set of keys vertically upward to his fraternity brother, who is in a window 4.00 m above. The brother’s outstretched hand catches the keys 1.50 s later.

(a) With what initial velocity were the keys thrown?
(b)What was the velocity of the keys just before they were caught?

Answer 1

Answer:10.02 m/s

4.68 m/s

Step-by-step explanation:

Given

height of building=4 m

time taken=1.5 s

(a)Let u be the initial velocity

using equation of motion

`s=ut+(gt^2)/(2)`

`4=u* 1.5-(9.81* 1.5^2)/(2)`

`8=u* 3-9.81* 2.25`

u=10.02 m/s

(b)Velocity of keys just before keys were caught

`v^2-u^2=2as`

`v^2=10.02^2-2(-9.81)\cdot 4`

`v^2=21.92`

`v=√(21.92)=4.68 m/s`