If a projectile is shot at an angle of (2pi)/32π3 and at a velocity of 6 m/s6ms, when will it reach its maximum height?

1 Answer
Dec 27, 2015

The projectile will reach its maximum height at 0.53 s

Explanation:

We consider the equation
s=vt-1/2at^2s=vt12at2 and v=v-atv=vat
we find the moment when the horizontal component of speed is 0 (the vertical component is v(x)=vsin(a))
so we have the equation v-at=0vat=0
a is equal to g, so t=v/g=0.53 m/s^2t=vg=0.53ms2