A ball is dropped from the top of building .The ball takes 0.5 sec to fall past the 3 m length of a window some distance from the top of building ,How fast was the ball going as it passes the top of the window ? How far is top of the window from the point at which ball was dropped ? ( take g = 9.8 m/s2 )
Let, ‘vt’ be the speed of the ball at the top point of the window, ‘vb’ be the speed at the bottom point of the window. The acceleration we know is ‘g’, g = 9.8 m/s2
The time taken to cross the window, t =0.5 s and the length of the window is S = 3 m.
Using,
S = ut + ½ at2
=> 3 = vt × t + ½ × g × t2
=> 3 = vt × 0.5 + ½ × 9.8 × 0.52
=> vt = 3.55 m/s
When the ball was just dropped its initial velocity was 0. So, the distance after which the velocity is vt = 3.55 m/s is (it is the distance from the point from which the ball is dropped to the top of the window),
v2 = u2 + 2as
=> 3.552 = 0 + 2 × 9.8 × s
=> s = 0.64 m