Answer to Question #93486 in Mechanics | Relativity for E

Question #93486
A baseball is dropped from the roof of a building. After some time, it flies past a window of height 2.20m. if the time it takes the baseball to fall past the window is 1/3s, how far above the window is the roof?
1
Expert's answer
2019-08-29T09:52:36-0400

Let vv be the velocity of ball just before it comes to the upper part of the window

Using 2nd equation of motion when the ball passing through window

s=(u×t)+(a2×t2)s=(u\times t)+(\frac{a}2\times t^2)

now

s=2.20ms=2.20m

t=13sec=\frac{1}{3}sec

a=g=9.8msec2a=g=9.8\frac{m}{{sec}^2}


2.20=(v×13)+(9.82×13×13)2.20=(v\times{\frac1 {3}})+(\frac{9.8}{2}\times\frac{1}{3}\times\frac{1}{3})


hence, v=4.95msecv=4.95\frac{m}{sec}

Using 3rd equation of motion from starting point to window

v2u2=2×a×sv^2-u^2=2\times a\times s

u=0u=0 as ball started from rest

v=4.95msecv=4.95\frac{m}{sec}

a=g=9.8msec2a=g=9.8\frac{m}{{sec}^2}

so,

4.952=2×9.8×s4.95^2=2\times 9.8\times s

s=1.25ms=1.25m

Roof is 1.25m1.25m  above the window


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