Aball is dropped from rest at point o. after falling for some time, it passes by a window of height 3.3 m and it does so in 0.27 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. calculate the speed va at which the ball passes the window’s top.

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

For the bottom of window (position B)

We also have

Solving

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

Speed at which the ball passes the window’s top = 10.89 m/s                

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