First slide

Vertical projection from top of a tower

Question

$\vec{u}\,=\,2\hat{i}$ A body is projected vertically upward with a velocity of 10 m/s from the top of a tower. At the same time a second body is projected vertically downward with a velocity of 5 m/s . Then velocity of the second body, relative to the first one after 3 second.will be
(3) 16 m/s

Moderate

A
10 m/s
B
18 m/s
C

15 m/s
D
12 m/s

Solution

Time taken by the first body to reach the highest point of its trajectory = $\frac{u}{g} = \frac{10}{10} s = 1 s$
Therefore after 3 s both of the bodies will be moving downward .
So $v_{2 / 1} = (5 + 10 \times 3) - (10 \times 2) m / s = 15 m / s$

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