# A bullet hit a target 301.5 m away. what maximum height above the muzzle did the bullet reach if it was shot at an angle of 25 degrees to the ground?

*print*Print*list*Cite

### 1 Answer

Let us say the velocity of the bullet is vm/s.

Using the equations of motion;

`rarr S = ut+1/2gt^2 `

`301.5 = vcos25t ------(1)`

`uarr S = ut+1/2g(t)^2`

`0 = vsin25t-1/2xx9.81xx(t)^2`

`vsin25 = 4.905t ----(2)`

(2)/(1)

`301.5/vsin25 = (vcos25t)/(4.905t)`

`301.5xx4.905 = v^2sin25cos25`

We know that;

`sin(2theta) = 2sinthetaxxcostheta`

`301.5xx4.905 = v^2sin25cos25`

`1478.8575 = v^2xx1/2xxsin50`

`v^2 = 1478.8575xx2/(sin50) `

`v^2 = 3861.0227`

`v = 62.137`

`uarr V^2 = U^2+2aS`

At the max height vertical component of v is 0.

`0 = (vsin25)-2xx9.81xxS`

`S = (vsin25)/(2xx9.81)`

`S = 35.15`

*So the bullet will reach a maximum height of 35.15m.*

**Sources:**