how do I solve the following problem?

Question

A gun is shot 2 m above the ground and exits the exits the gun at 370 m/s at an angle of 6 degrees above the ground. How far will the bullet travel before returning to the 2 meters above the ground?

Answer 1

Hi Sherisse

 

The horizontal range at 2m above the fround i given as 

R= V² Sin2theta/g

 

= (370*370)( Sin 12)/9.8

= 2904.4 m

It will go further as it fall down another 2m with horizontal velocity of V cos theta  and vertically downward velocity of V sin theta

 

time taken to fall vertically is given by

d= ut + 1/2 gt²

u = initial vertical velocity = V Sin theta

 

2= (370 *t *(sin 6) +1/2 gt²

 

Solve this equation for t 

once you get " t "

 

Then use Velocity = distance /time 

V Cos(theta ) = X /t

or 370 Cos 6 = X/t

From here get " X "

 

The total horizontal distance covered =( 2904.4 + X)

Use Sin 6= 0.104 and Cos 6= 0.994