Question

A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outstretched hand catches the keys 1.60 s later. (Take upward as the positive direction. Indicate the direction with the sign of your answer.)With what initial velocity were the keys thrown?

Answer 1

Answer:

$v_{i}=10.10 m/s$

Explanation:

The equation of the position is:

$y=y_{i}+v_{i}t-0.5gt^{2}$

Where:

v(i) is the initial velocity

The initial position y(i) will be zero and the final position y = 3.60 m.

So, we just need to solve this equation for v(i).

$v_{i}=\frac{y+0.5gt^{2}}{t}$

$v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}$

$v_{i}=10.10 m/s$

Therefore, the initial velocity is 10.10 m/s upwards.

I hope it helps you!