Q: A ball is dropped from top of a building. The ball takes 0.5 s to fall past the 3 m length of a window some distance from top of building. With what speed does the ball pass the top of window ?

(a) 6 m/s

(b) 12 m/s

(c) 7 m/s

(d) 3.5 m/s

Ans:(d)

Sol: Let x be the distance of top of window from top of the building, and let the ball take t sec to reach there.

Applying ,

$\displaystyle s = u t + \frac{1}{2} a t^2 $

$ \displaystyle x = 0 + \frac{1}{2} (10) t^2 $

x = 5 t ^{2} …(i)

Now ,

$ \displaystyle x + 3 = 0 + \frac{1}{2} (10) (t + 0.5)^2 $

$\displaystyle x + 3 = 5(t^2 + t + 1/4) $ …(ii)

Subtract (i) from (ii)

3=5(1/4+t)

3 =5/4 + 5t

3-5/4 = 5t

7/4 = 5t

or t = 7/20 sec

From, v = u + at

v = 0 + 10× 7/20

= 3.5 m/s