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Rus_ich [418]
3 years ago
15

A student throws a set of keys vertically upward to his fraternity brother, who is in a window 4.00 m above. The brother’s outst

retched hand catches the keys 1.50 s later.
(a) With what initial velocity were the keys thrown?
(b)What was the velocity of the keys just before they were caught?
Physics
1 answer:
MakcuM [25]3 years ago
3 0

Answer:10.02 m/s

4.68 m/s

Explanation:

Given

height of building=4 m

time taken=1.5 s

(a)Let u be the initial velocity

using equation of motion

s=ut+\frac{gt^2}{2}

4=u\times 1.5-\frac{9.81\times 1.5^2}{2}

8=u\times 3-9.81\times 2.25

u=10.02 m/s

(b)Velocity of keys just before keys were caught

v^2-u^2=2as

v^2=10.02^2-2(-9.81)\cdot 4

v^2=21.92

v=\sqrt{21.92}=4.68 m/s

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