Question #f551a

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Answer 1 · 2017-04-11T14:05:47

The horizontal range is $= \frac{4 u^{2}}{5 g}$ $=(4u^2)/(5g)$

Resolving in the vertical direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

initial velocity is $u_{y} = u sin θ$ $u_y=usintheta$

Acceleration is $a = - g$ $a=-g$

At the maximum height, $v = 0$ $v=0$

We apply the equation of motion

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

to calculate the greatest height

$0 = u^{2} {sin}^{2} θ - 2 \cdot g \cdot h$ $0=u^2sin^2theta-2*g*h$

$h = \frac{u^{2} {sin}^{2} θ}{2 g}$ $h=(u^2sin^2theta)/(2g)$

We apply the equation of motion

$v = u + a t$ $v=u+at$

to calculate the time to reach the greatest height

$0 = u sin θ - g \cdot t$ $0=usintheta-g*t$

$t = \frac{u}{g} \cdot sin θ$ $t=u/g*sintheta$

Resolving in the horizontal direction $\to^{+}$ $rarr^+$

We apply the equation of motion

$s = u_{x} \cdot t$ $s=u_x*t$

$= 2 u cos θ \cdot \frac{u}{g} \cdot sin θ$ $=2ucostheta*u/g*sintheta$

$= 2 \frac{u^{2}}{g} cos θ sin θ$ $=2u^2/gcosthetasintheta$

It is given that

$s = 2 h$ $s=2h$

Therefore,

$2 \frac{u^{2}}{g} cos θ sin θ = 2 \cdot \frac{u^{2} {sin}^{2} θ}{2 g}$ $2u^2/gcosthetasintheta=2*(u^2sin^2theta)/(2g)$

$cos θ sin θ = \frac{{sin}^{2} θ}{2}$ $costhetasintheta=sin^2theta/2$

$tan θ = 2$ $tantheta=2$

$θ = {63.4}^{\circ}$ $theta=63.4^@$

So,

The horizontal range is

$s = 2 \frac{u^{2}}{g} cos θ sin θ$ $s=2u^2/gcosthetasintheta$

$= 2 \frac{u^{2}}{g} cos ({63.4}^{\circ}) sin ({63.4}^{\circ})$ $=2u^2/gcos(63.4^@)sin(63.4^@)$

$= 2 \frac{u^{2}}{g} \cdot \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}}$ $=2u^2/g*1/sqrt5*2/sqrt5$

$s = \frac{4 u^{2}}{5 g}$ $s=(4u^2)/(5g)$