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

Explanation:

Given

height of building=4 m

time taken=1.5 s

(a)Let u be the initial velocity

using equation of motion

$s=ut+\frac{gt^2}{2}$

$4=u\times 1.5-\frac{9.81\times 1.5^2}{2}$

$8=u\times 3-9.81\times 2.25$

u=10.02 m/s

(b)Velocity of keys just before keys were caught

$v^2-u^2=2as$

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

$v^2=21.92$

$v=\sqrt{21.92}=4.68 m/s$