A ball is dropped from rest at point O (height unknown). After falling for some

Question

A ball is dropped from rest at point O
(height unknown). After falling for some
time, it passes by a window of height 2.8 m
and it does so during time tAB = 0.44 s.
The acceleration of gravity is 9.8 m/s2 .

The ball accelerates all the way down; let
VA be its speed as it passes the window’s top
A and VB its speed as it passes the window’s
bottom B.

1- How much did the ball speed up as it passed the window; i.e., calculate Vdown = vBvA ? Answer in units of m/s.

2- Calculate the speed VA at which the ball passes the window’s top. Answer in units of m/s.

A ball is dropped from rest at point O (height unknown). After falling for some time, it passes by a window of height 2.8 m and it does so during time tAB = 0.44 s. The acceleration of gravity is 9.8 m/s2 The ball accelerates all the way down; let VA be its speed as it passes the window?s top A and VB its speed as it passes the window?s bottom B. 1- How much did the ball speed up as it passed the window; i.e., calculate Vdown = vB?vA ? Answer in units of m/s. 2- Calculate the speed VA at which the ball passes the window?s top. Answer in units of m/s.

Explanation / Answer

Since it accelerates at the rate of 9.8 m/sec^2, in 0.44 sec it will increase in velocity 9.8*0.44= 4.312 m/sec To do 2 we solve the equation 2.8= Vo (0.44) +4.9 (0.44)^2 2.8= 0.44 Vo +0.94864 0.44 Vo= 1.85136 Vo= 4.208 m/sec

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