# If a projectile is shot at an angle of pi/6 and at a velocity of 15 m/s, when will it reach its maximum height??

Jan 3, 2016

Assuming $g = 9.8 m {s}^{- 2}$ it will reach its maximum height after $0.765$ seconds.

#### Explanation:

The vertical component of the velocity is $15 \cdot \sin \left(\frac{\pi}{6}\right) = 7.5 m {s}^{- 1}$

The gravitational acceleration is $9.8 m {s}^{- 2}$.

So it will take $\frac{7.5}{9.8} \approx 0.765$ seconds before the vertical component of the velocity is zero.

Jan 4, 2016

$0.76 s$

#### Explanation:

I always find a diagram helpful:

I'm going to make the assumption for this question that we can ignore air resistance.
I am also defining right and up as positive directions.
This allows us to use the constant acceleration equations.

It should also be noted that at the maximum height, vertical velocity is zero.

This equation seems ideal for this problem, as it contains all the variables that we have values for and the one we wish to solve.

${v}_{y} = {u}_{y} + {a}_{y} t$

Let's see what values we have:

${v}_{y} = 0$
${u}_{y} = 15 \sin \left(\frac{\pi}{6}\right) = 7.5$
${a}_{y} = - g = - 9.81$
t=?

Plugging them in:

$0 = 7.5 - 9.81 t$

$t = \frac{7.5}{9.81}$

$t = 0.76452 \ldots s$

$t = 0.76 s$ (2sf)