Question

A projectile is fired at ????0=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, ???? ?

Answer 1

Answer:

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 $g=9.81m/sec^2$

Horizontal range is given by $R=\frac{u^2sin2\Theta }{g}=\frac{355^2sin(2\times 68.4^{\circ})}{9.81}=8794.09m$

It is given that range is shorten by 35.1%

So range will be $8794.09-(8794.09\times \frac{35.1}{100})=5707.364m$