Question

Physics Projectile Sum Please Help Me If you know
A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) What is the size of the angle θ?
b) At what initial velocity was the projectile launched?

Answer 1

Answer:

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NIVEDITA

Explanation:

Use the following equation of motion since we are interested in finding velocity:

=+vf=vi+at

Since we have motion in both the x-direction and the y-direction, write the above Newton’s equation in the y-direction to determine the vertical velocity. Use subscripts to help:

()=()+(vf)y=(vi)y+ayt  —————— eqn 1

Always watch your sign convention when using Newton’s equations of motion. I always assume positive = up and negative = down, so the acceleration of gravity is *always* negative:

=−9.812ay=−9.81ms2

and the initial velocity is positive since it is upward:

Equation 1 becomes:

()=+3060+(−9.812)(5)(vf)y=+30sin60+(−9.81ms2)(5sec)

()=−23.07(vf)y=−23.07ms

The negative answer indicates it is moving downward.

The projectile also has horizontal velocity. Use the same Newton’s equation of motion, except write the equation in the x-direction using subscripts:

()=()+(vf)x=(vi)x+axt

Since we are ignoring air resistance,  =0ax=0

∴()=()∴(vf)x=(vi)x

()=+3060=15(vf)x=+30cos60=15ms

Using Pythagoras:  =()2+()2‾‾‾‾‾‾‾‾‾‾‾‾√=27.5v=(vx)2+(vy)2=27.5ms

in a direction down and to the right at an angle  =arctan()=57∘θ=arctan⁡(vyvx)=57∘  below the horizontal