Question

A student throws a set of keys vertically upward to his fraternity brother who is in a window 3.60 m above. The brothers outstretched hand catches the keys 1.45 seconds later. With what initial velocity where the keys thrown? What was the velocity of the keys just before they were caught?

Answer 1

The initial velocity where the keys thrown is 9.59 m/s

The velocity of the keys just before they were caught is -4.62 m/s

Explanation:

Given:

Height (h) = 3.6 m

Time (t) = 1.45 sec

We can calculate the initial velocity using the following formula:

`h=ut+(1)/(2)at^2`

We replace and solve for u (initial velocity)

`\begin{gathered} 3.6=u\cdot1.45+(1)/(2)\cdot(-9.8)\cdot(1.45)^2 \\ \\ 3.6=1.45u-10.3 \\ \\ 1.45u=3.6+10.3 \\ \\ u=(13.9)/(1.45) \\ \\ u=9.59\text{ m/s} \end{gathered}`

So, the initial velocity where the keys thrown is 9.59 m/s

Now, we calculate the final velocity using the following formula:

`\begin{gathered} v=u+at \\ \\ \text{ we replacing} \\ \\ v=9.59+(-9.8)(1.45) \\ \\ v=9.59-14.21 \\ \\ v=-4.62\text{ m/s} \end{gathered}`

Therefore, the velocity of the keys just before they were caught is -4.62 m/s