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A student throws a set of keys vertically upward to his fraternity brother, who is in a window 4.35 m above. The brother's outstretched hand catches the keys 1.20 s later. (Take upward as the positive direction. Indicate the direction with the sign of your answer.) (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught?

Respuesta :

Answer:9.51 m/s

Explanation:

Given

Window is 4.35 m above ground

time taken=1.2 s

Let u be the initial velocity of keys

[tex]s=ut+\frac{gt^2}{2}[/tex]

where s=distance traveled

u=initial velocity

t=time taken

g=acceleration due to gravity

[tex]4.35=u\times 1.2+\frac{(-9.8)1.2^2}{2}[/tex]

[tex]u=\frac{11.4132}{1.2}=9.511 m/s[/tex]

(b)velocity of keys just before they were caught

v=u+at

[tex]v=9.511-9.8\times 1.2[/tex]

v=-2.26 m/s

Here negative sign indicates downward motion of keys

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