Asked Dec 16, 2019

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
outstretched 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?


Expert Answer

Step 1

(a)Write the kinematic equation of motion.


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S= 4.00m. t=1.50 s S= ut- La=g=-9.8m/s 4.00m = 1(1.50 s) +(-9.8 m/s')(1.50 s) u= 10m/s

Step 2

(b)The time of ascent of the key is.


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10m/s 9.8 m/s =1.02 s

Step 3

The time of falling from ...


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t=1.50s-1.02s =0.48s


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