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