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

`v_(i)=10.10 m/s`

Step-by-step 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)=(y+0.5gt^(2))/(t)`

`v_(i)=(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!