First slide

Vertical projection from top of a tower

Question

A ball is thrown vertically upwards with a speed of 10 m/s and another ball is thrown vertically downwards with the same speed simultaneously. The time difference between them in reaching the ground )(in s) is: $(g = 10 m / s^{2})$

Easy

A

12
B

6
C

2
D

1

Solution

For both of them displacements are same. $h = - u t_{1} + \frac{1}{2} g {t_{1}}^{2} a n d h = u t_{2} + \frac{1}{2} g t_{2}^{2} s o l v i n g w e g e t u = \frac{1}{2} g (t_{1} - t_{2})$

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