Question

If a projectile is shot at an angle of `pi/2` and at a velocity of `64 m/s`, when will it reach its maximum height??

Answer 1

It reaches its highest point `6.5 s` after launch.

Explanation:

`pi/2` is straight up so we do not need to calculate a vertical component of the initial velocity. We can use the kinematic formula

`v = u + a*t`

where `v = 0` because it stops at maximum height,
`u = 64 m/s`,
and `a = -9.8 m/s^2` which is the acceleration due to gravity, g. The value of a gets a minus sign because it is in the opposite direction of the initial velocity.

`v = u + a*t`

`0 = 64 m/s - 9.8 m/s^2*t`

Solving for t,

`t = (64 cancel(m)/s)/(9.8 cancel(m)/s^2)`
(hmmm - I struggled with the formatter here. I wanted to strike out the power of 2 on the s, leaving just s. Would not work for me.)

`t = 64 /9.8 s = 6.53 s ~= 6.5 s`

I hope this helps,
Steve