Question

Question #f551a

Answer 1

The horizontal range is `=(4u^2)/(5g)`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=usintheta`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v^2=u^2+2as`

to calculate the greatest height

`0=u^2sin^2theta-2*g*h`

`h=(u^2sin^2theta)/(2g)`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=usintheta-g*t`

`t=u/g*sintheta`

Resolving in the horizontal direction `rarr^+`

We apply the equation of motion

`s=u_x*t`

`=2ucostheta*u/g*sintheta`

`=2u^2/gcosthetasintheta`

It is given that

`s=2h`

Therefore,

`2u^2/gcosthetasintheta=2*(u^2sin^2theta)/(2g)`

`costhetasintheta=sin^2theta/2`

`tantheta=2`

`theta=63.4^@`

So,

The horizontal range is

`s=2u^2/gcosthetasintheta`

`=2u^2/gcos(63.4^@)sin(63.4^@)`

`=2u^2/g*1/sqrt5*2/sqrt5`

`s=(4u^2)/(5g)`