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

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Answer 1 · 2016-04-18T00:21:04

Time $t = \frac{3 \cdot sin (\frac{π}{8})}{9.8} = 0.1171479895$ $t=(3*sin(pi/8))/9.8=0.1171479895" "$ second

For projectiles fired at angle $θ$ $theta$ with initial velocity $v_{0}$ $v_0$

$y = v_{0} sin θ \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $y=v_0 sin theta* t+1/2*g*t^2$

We are after the maximum height $y_max$ .
We differentiate $y$ with respect to $t$ then equate $dy/dt=0$ because we maximizing height $y$

$y=v_0 sin theta* t+1/2*g*t^2$

$dy/dt=v_0 sin theta *dt/dt+1/2*g*2*t*dt/dt$

$dy/dt=v_0 sin theta *1+1/2*g*2*t*1$

$dy/dt=v_0 sin theta+g*t$

$dy/dt=0$

$v_0 sin theta+g*t=0$

Solving for t

$t=(-v_0 sin theta)/g=(-3*sin 22.5^@)/(-9.8)=0.1171479895" "$ second

God bless...I hope the explanation is useful.

If a projectile is shot at an angle of (pi)/8π8(pi)/8 and at a velocity of 3 m/s3ms3 m/s, when will it reach its maximum height?