Question

A projectile is fired at v0 = 420.0 m/s at an angle of θ = 67.2o 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, R?

Answer 1

First calculate time to full height.
Initial Vertical velocity will be Vo*Sin67.2= 387.2 m/sec
Vf=Vi- g*t.
0=387.2-9.8*t
Time to full height t= 387.2/9.8= 39.46sec.
Time for hor. flight =2*39.46= 78.93 S
Hor velocity is constant (neglecting air resistance) at VCos71.7= 127.17 m/s
Total distance R= 420*cos(67.2)* 78.93= 12846.4 meters.
Compensating for air resistance @ 35.1% R= 12846.4-4509= 8337.4 meters