Projectile Motion

Question

A projectile is fired with a velocity u at an angle θ with the horizontal. At the highest point of its trajectory, the projectile explodes into two fragments of equal mass. One fragment falls vertically downwards with zero initial velocity. The distance of the other fragment from the point of projection when it strikes the ground is:

RankGuru · Accepted Answer

1. **At the highest point**: The projectile has only a horizontal component of velocity, $u_x = u \cos	heta$. The vertical velocity is zero.
2. **Momentum Conservation (Horizontal)**: Before the explosion, momentum is $m(u \cos	heta)$. After explosion, it is $\frac{m}{2}(0) + \frac{m}{2}v'$. 
   $$m u \cos	heta = \frac{m}{2}v' \implies v' = 2u \cos	heta$$
3. **Time to reach ground**: Since the fragments are at height $H$ and have zero vertical velocity initially, the time taken for the second fragment to hit the ground is the same as the first: $t = \frac{u \sin	heta}{g}$.
4. **Distance from highest point**: The second fragment covers a horizontal distance $X' = v' \cdot t = (2u \cos	heta)\left(\frac{u \sin	heta}{g}ight) = \frac{2u^2 \sin	heta \cos	heta}{g} = R$.
5. **Total Distance**: Since the explosion happened at $R/2$, the total distance from the origin is $\frac{R}{2} + R = 1.5R$.

Rank Guru

Projectile Motion

Select the correct option:

Solution

🔒 Solution Hidden from View

More projectile motion Practice Questions

A projectile is launched with speed 20 m/s at 30° above horizontal. Ignore air resistance. What is i...

A projectile is fired at an angle of 30° with the horizontal with an initial velocity of 40 m/s. Wha...

A projectile is fired at 20 m/s at 30° above horizontal. Time of flight (g=10 m/s²) is:

Maximum height reached by the projectile of KN-003 is:

Range is maximum when launch angle is:

About This Question