Question

A bullet is shot at an angle of 32° above the horizontal on a level surface. It travels in the air for 6.4 seconds before it strikes the ground 92m from the shooter. What was the maximum height reached by the bullet?

Answer 1

H = 4.12 m

Step-by-step explanation:

As we know that horizontal range is the distance moved in horizontal direction

Since horizontal direction has no acceleration

so here we have

`Range = v_x T`

here we know that

`v_x = vcos32`

so from above formula

`92 = (vcos32)(6.4)`

`v = 16.95 m/s`

now we have maximum height is given as

`H = ((vsin32)^2)/(2g)`

`H = ((16.95 sin32)^2)/(2(9.8))`

`H = 4.12 m`