Question

A shell fired from a gun has a horizontal range of 1200 m and a flight time of 10 s. What is the magnitude and direction of the velocity of projection?

Answer 1

`|v| = 129.64 m/s`
`theta = 22.23°`

Explanation:

`x = v_{0x} * t = v_{0x} * 10 = 1200`
`=> v_{0x} = 120 m/s`
`v_y = v_{0y} - g*t`
`"At the highest point "v_y = 0" and "t = 10/2 = 5 s`
`=> 0 = v_{0y} - 9.81*5`
`=> v_{0y} = 49.05 m/s`

`=> |v| = sqrt(v_{0x}^2 + v_{0y}^2) = 129.64 m/s`
`theta = arctan(v_{0y}/v_{0x}) = 22.23°`
`"So the direction is 22.23° above the horizontal ground."`

Answer 2

The velocity of projection is `=129.6ms^-1` and the angle is `=22.2^@`

Explanation:

The range is

`r=v^2sin(2theta)/g`

The time of flight is

`t=2vsintheta/g`

Therefore,

`v^2sin(2theta)/g=1200`................`(1)`

and

`2vsintheta/g=10`........................`(2)`

So,

Equation `(1)` divided by `(2)` is

`vcostheta=120`

The acceleration due to gravity is `g=9.8ms^-2`

From `(2)`

`vsintheta=5g=49`

`=>`, `(vsintheta)/(vcostheta)=49/120`

`tantheta=0.4083`

`theta=arctan(0.4083)=22.2^@`

`v=120/costheta=120/cos(22.2^@)=129.6ms^-1`