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

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 10 ~s

A projectile moving vertically upwards with a velocity of 200 ~ms^ -1 breaks into two equal parts at a height of 490 ~m

Momentum before explosion = Momentum after explosion

\begin{gathered}m \times200 \hat{j}=\frac{m}{2} \times400 \hat{j}+\frac{m}{2}v \\ =\frac{m}{2}(400 \hat{j}+v) \\ \Rightarrow \quad 400 \hat{j}-400 \hat{j}=v \\ \therefore \quad v=0 \end{gathered} 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

\begin{aligned}& 490=0+\frac{1}{2} \times9.8 \timest^2 \\ & \Rightarrow \quad t²=\frac{980}{9.8}=100 \\ & \therefore \quad t=\sqrt{100}=10 \mathrm{sec} \end{aligned}

A projectile moving vertically upwards with a velocity of 200 ~ms^...

Options

Correct answer

Step-by-step solution

More from Center of Mass, Momentum and Collision