Question

For the same two projectiles, after $3s$ from the initial launch, what will be the difference between the two projectiles' speeds?
The previous problem's text:
A projectile is fired $30.0$ degrees above the horizontal with speed $|v_1|=40m/s$ and a second one $60.0$ degrees above the horizontal with speed $|v_2|=30m/S$ simultaneously.
After 2 seconds, how far apart will the two projectiles be? Assume no air resistance and that they are fired from the same spot, and that each moves independent of the outer projectile. Take $g=9.81m/s^2$.

• A. $10m/s$
• B. $15.4m/s$
• C. $20.5m/s$
• D. $30.0m/s$
• E. $35.9m/s$

Answer 1

The correct option is C $20.5m/s$

The horizontal component of velocity of both projectile remains the same,

$u_{x_1}=40m/s\times cos{30}^{\circ }$

$u_{x_2}=30m/s\times cos{60}^{\circ }$

The vertical components of velocity changes under gravity.

$v_{y_1}=u_{y_1}-gt$

$=(40\times sin{30}^{\circ })-(9.81\times 3)=-9.43m/s$

$v_{y_2}=u_{y_2}-gt$

$=-3.45m/s$

Hence difference in speeds=$|\sqrt{u_{x_1}^2+v_{y_1}^2}-\sqrt{u_{x_2}^2-v_{y_2}^2}|$

$=20.5m/s$