# If a projectile is shot at an angle of pi/6 and at a velocity of 5 m/s, when will it reach its maximum height??

Mar 11, 2016

$0.25 \text{s}$

#### Explanation:

The initial upwards component of the velocity is

(5 "m/s") * sin(pi/6) = 2.5 "m/s"

As the projectile is in free fall, the object is decelerating at $g = 9.81 {\text{m/s}}^{2}$.

At the highest point, the vertical component of the projectile's velocity is zero. Think about why this is true. If the projectile still has velocity in the upwards direction, it can still go further up. If the projectile has velocity in the downwards direction, it must have descended from a place that is higher that its current position.

An object traveling under constant acceleration $a$, with initial velocity $u$, has velocity $v$ at a time $t$ given by

$v = u + a t$

We have

• $u = 2.5 \text{m/s}$
• $a = - 9.81 {\text{m/s}}^{2}$

Note: $a$ has to be of the opposite sign of $u$, as the acceleration (downwards) is in the opposite direction of the initial velocity (upwards).

So, to solve for the maximum height, we need to solve for $v = 0$.

$v = u + a t = 0$

$u + a t = 0$

2.5 "m/s" + (-9.81 "m/s"^2) t = 0

Solving for $t$ directly yields $t = 0.25 \text{s}$.

That's it.

Now, you may be wondering about the horizontal component of the projectile's velocity. Since there is no horizontal net force, the horizontal velocity remains at a value of $5 \text{m/s} \times \cos \left(\frac{\pi}{6}\right)$.