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

Question

a) A ball is dropped from rest at point O (height unknown). After falling for some time, it passes by a window of height 2.4 m and it does so in 0.37 s. 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. How much did the ball speed up as it passed the window; i.e., calculate vdown = vB vA ? The acceleration of gravity is 9.8 m/s^2 . Answer in units of m/s. Show work.

b)   Calculate the speed vA at which the ball passes the window’s top. Answer in units of m/s. Show work.

Explanation / Answer

I'll give you the top & bottom speeds, you can sort from there anything else.

In this type of exercise, first find the average velocity.

this is done by:

h/t = 2.4/0.37 = 6.486 m/s

g = 9.8 m/s^2

Top velocity = average speed - ((t/2) x g)

= 6.486 - ((0.37/2) x 9.8) = 4.673m/s


Bottom velocity = average speed + ((t/2) x g)

= 6.486 + ((0.37/2) x 9.8) = 8.299 m/s

change in velocity =VB-VA

                           =8.299-4.673=3.626m/s

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