# 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??

Mar 10, 2018

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

#### Explanation:

$\frac{\pi}{2}$ is straight up so we do not need to calculate a vertical component of the initial velocity. We can use the kinematic formula

$v = u + a \cdot t$

where $v = 0$ because it stops at maximum height,
$u = 64 \frac{m}{s}$,
and $a = - 9.8 \frac{m}{s} ^ 2$ 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.

$v = u + a \cdot t$

$0 = 64 \frac{m}{s} - 9.8 \frac{m}{s} ^ 2 \cdot t$

Solving for t,

$t = \frac{64 \frac{\cancel{m}}{s}}{9.8 \frac{\cancel{m}}{s} ^ 2}$
(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.)

$t = \frac{64}{9.8} s = 6.53 s \cong 6.5 s$

I hope this helps,
Steve