Question

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. $11.96m$
• B. $28.11m$
• C. $39.28m$
• D. $41.06m$
• E. $41.99m$

Answer 1

Answer: (D)

$41.06m$

In projectile motion the displacement in vertical direction is given by ,

$y=\left(v\sin \theta \right)t-1/2g{t}^2$ ,

where $v=$initial velocity ,

$\theta =$angle of projectile with horizontal

for first projectile , $\theta ={30}^o,v=40m/s,t=2s$

$y_1=40\sin 30\times 2-1/2\times 9.81\times 2^2$

or $y_1=40-19.62=20.38m$

for second projectile , $\theta ={60}^o,v=30m/s,t=2s$

$y_2=30\sin 60\times 2-1/2\times 9.81\times 2^2$

or $y_2=51.96-19.62=32.34m$

vertical distance between two projectile $y_2-y_1=32.34-20.38=11.96m$ ,

horizontal distance will also be $x=v\cos \theta t$. ,

for first projectile $x_1=40\cos 30\times 2=69.28m$

for second projectile $x_2=30\cos 60\times 2=30m$

therefore horizontal distance $x=69.28-30=39.28m$

now by pythagoras theorem , the hypotenuse of right angled triangle whose two sides are $11.96m$ and $69.28m$ will be the distance between two projectiles ,

distance $=\sqrt{{11.96}^2+{69.28}^2}=41.06m$