Question #f551a

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Answer 1 · 2017-04-11T14:05:47

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

Resolving in the vertical direction $uarr^+$

initial velocity is $u_y=usintheta$

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

$s=2u^2/gcosthetasintheta$

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

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

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