If a projectile is shot at an angle of $\frac{3 π}{8}$ $(3pi)/8$ and at a velocity of $2 \frac{m}{s}$ $2m/s$ , when will it reach its maximum height?

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Answer 1 · 2018-06-26T06:13:45

The time is $= 0.19 s$ $=0.19s$

The initial speed is $u = 2 m s^{- 1}$ $u=2ms^-1$

The angle is $θ = \frac{3}{8} π r a d$ $theta=3/8pirad$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The equation of the trajectory of the projectile is

$y = x tan θ - \frac{g x^{2}}{2 u^{2} {cos}^{2} θ}$ $y=xtantheta-(gx^2)/(2u^2cos^2theta)$ .............. $(1)$ $(1)$

At the maximum height

$\frac{d y}{d x} = 0$ $dy/dx=0$

$\Rightarrow$ $=>$ , $\frac{d y}{d x} = tan θ - \frac{2 g x}{2 u^{2} cos θ}$ $dy/dx=tantheta-(2gx)/(2u^2costheta)$

$= tan θ - \frac{g x}{u^{2} {cos}^{2} θ}$ $=tantheta-(gx)/(u^2cos^2theta)$

Therefore,

$tan θ - \frac{g x}{u^{2} {cos}^{2} θ} = 0$ $tantheta-(gx)/(u^2cos^2theta)=0$

$x = \frac{u^{2} tan θ {cos}^{2} θ}{g} = u^{2} \frac{sin θ cos θ}{g}$ $x=(u^2tanthetacos^2theta)/(g)=u^2(sinthetacostheta)/g$

The constant horizontal velocity is

$v_{x} = u cos θ$ $v_x=ucostheta$

Therefore,

The time to reach the greatest height is

$t = \frac{x}{v_{x}} = (u^{2} \frac{sin θ cos θ}{g}) \cdot \frac{1}{u cos θ}$ $t=x/v_x=(u^2(sinthetacostheta)/g)*1/(ucostheta)$

$= \frac{u sin θ}{g}$ $=(usintheta)/g$

$= 2 \cdot \frac{sin (\frac{3}{8} π)}{9.8}$ $=2*sin(3/8pi)/9.8$

$= 0.19 s$ $=0.19s$

If a projectile is shot at an angle of (3pi)/83π8(3pi)/8 and at a velocity of 2m/s2ms2m/s, when will it reach its maximum height?