Question

If a projectile is shot at an angle of `(2pi)/3` and at a velocity of `6 m/s`, when will it reach its maximum height?

Answer 1

The projectile will reach its maximum height at 0.53 s

Explanation:

We consider the equation
`s=vt-1/2at^2` and `v=v-at`
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=0`
a is equal to g, so `t=v/g=0.53 m/s^2`