Aprojectile is fired at =355.0 m/s at an angle of theta=68.4∘ , with respect to the horizontal. assume that air friction will shorten the range by 35.1% . how far will the projectile travel in the horizontal direction, ?

Range will be 5707.364 m      

Explanation:

We have given initial velocity of projectile u = 355 m/sec

Angle of projection = 68.4°

Acceleration due to gravity

Horizontal range is given by

It is given that range is shorten by 35.1%

So range will be

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

read more on -

grav potential energy