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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?

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whitneytr12

Answer:

[tex]v_{i}=10.10 m/s[/tex]

Explanation:

The equation of the position is:

[tex]y=y_{i}+v_{i}t-0.5gt^{2}[/tex]

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).

[tex]v_{i}=\frac{y+0.5gt^{2}}{t}[/tex]

[tex]v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}[/tex]

[tex]v_{i}=10.10 m/s[/tex]

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

I hope it helps you!