A projectile is moving at 20 m / s at its highest point, where it breaks into equal parts due to an internal explosion. One part moves vertically up at 30 m / s with respect to he ground. Then the other part will move at:

Question

A projectile is moving at 20 m / s at its highest point, where it breaks into equal parts due to an internal explosion. One part moves vertically up at 30 m / s with respect to he ground. Then the other part will move at: (A) 10√13m/s (B) 20 m/s (C) 30 m/s (D) 50 m/s

Tardigrade Admin · Accepted Answer

Using momentum conservation, beginarrayl m × 20 hat i = m / 2 × 30 hat j + m / 2 × x x =40 hat i -30 hat j ⇒ | x |=50 endarray

Q. A projectile is moving at $20 m / s$ at its highest point, where it breaks into equal parts due to an internal explosion. One part moves vertically up at $30 m / s$ with respect to he ground. Then the other part will move at:

$1013 m / s$

$20 m / s$

$30 m / s$

$50 m / s$

Using momentum conservation,
$m \times 20 \hat{i} = m /2 \times 30 \hat{j} + m /2 \times x x = 40 \hat{i} - 30 \hat{j} \Rightarrow ∣ x ∣ = 50$

Q. A projectile is moving at 20m/s at its highest point, where it breaks into equal parts due to an internal explosion. One part moves vertically up at 30m/s with respect to he ground. Then the other part will move at:

1013​m/s

20m/s

30m/s

50m/s