A projectile moving vertically upwards with a velocity of 200 ms − 1 breaks into two equal parts at a height of 490 m . One part starts moving vertically upwards with a velocity of 400 ms − 1 . How mu

Question

A projectile moving vertically upwards with a velocity of 200 ms − 1 breaks into two equal parts at a height of 490 m . One part starts moving vertically upwards with a velocity of 400 ms − 1 . How much time it will take, after the break up with the other part to hit the ground?

TestHub · Accepted Answer

Momentum before explosion = Momentum after explosion $m \times 200 \hat{j} = \frac{m}{2} \times 400 \hat{j} + \frac{m}{2} v = \frac{m}{2} (400 \hat{j} + v) \Rightarrow 400 \hat{j} - 400 \hat{j} = v ∴ v = 0$ i.e., the velocity of the other part of the mass, $v = 0$ Let time taken to reach the earth by this part be $t$ Applying formula, $h = u t + \frac{1}{2} g t^{2}$ $490 = 0 + \frac{1}{2} \times 9.8 \times t^{2} \Rightarrow t^{2} = \frac{980}{9.8} = 100 ∴ t = 100 = 10 sec$

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