If a projectile is shot at an angle of $pi/3$ and at a velocity of $32 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-05-24T07:25:51

at time $t=2.827838053" "$ seconds

The equation for the vertical displacement for projectiles

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

The equation for the horizontal displacement for projectiles is

$x=v_0 cos theta *t$
Maximizing the vertical displacement by obtaining the first derivative of y with respect to time t

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

set $dy/dt=0$

$v_0 sin theta+1/2*g*2*t=0$

$t=(-v_0*sin theta)/g$

$t=(-32 sin ((pi)/3))/(-9.8)$

$t=(80sqrt3)/49$

$t=2.827838053" "$ seconds

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

If a projectile is shot at an angle of pi/3pi/3 and at a velocity of 32 m/s32 m/s, when will it reach its maximum height??