If a projectile is shot at a velocity of 6 m/s and an angle of pi/6, how far will the projectile travel before landing?

1 Answer
Apr 21, 2017

The distance is =1.02m

Explanation:

Resolving in the vertical direction uarr^+

initial velocity is u_y=vsintheta=6*sin(1/6pi)

Acceleration is a=-g

At the maximum height, v=0

We apply the equation of motion

v=u+at

to calculate the time to reach the greatest height

0=6sin(1/6pi)-g*t

t=6/g*sin(1/6pi)

=0.306s

Resolving in the horizontal direction rarr^+

To find the distance where the projectile will land, we apply the equation of motion

s=u_x*2t

=6cos(1/6pi)*0.106*2

=1.02m