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

Range will be 5707.364 m

Step-by-step 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=(u^2sin2\Theta )/(g)=(355^2sin(2* 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* (35.1)/(100))=5707.364m`