# A projectile is shot at a velocity of 37 m/s and an angle of pi/12 . What is the projectile's maximum height?

Dec 3, 2016

The maximum height is $= 4.59 m$

#### Explanation:

The maximum height is obtained from the formula

${v}^{2} = {u}^{2} - 2 g h$

The vertical component of the velocity is

$u = {u}_{0} \sin \theta$

At the maximum height,

$v = 0$

So,

$2 g h = {u}_{0}^{2} {\sin}^{2} \theta$

$h = \frac{{u}_{0}^{2} {\sin}^{2} \theta}{2 g}$

$h = {\left(37 \sin \left(\frac{\pi}{12}\right)\right)}^{2} / \left(2 \cdot 10\right)$

$= 4.59 m$

Dec 3, 2016

The maximum height of the projectile is $4.7 m$.

#### Explanation:

To find the projectile's maximum altitude, or maximum height, we can make use of kinematic equations. Namely, (v_f)_y^2=(v_i)_y^2+2a_yΔy will do nicely, as it won't require us to calculate the flight time of the projectile fist.

Because the projectile is launched at an angle, it is necessary to break up the given initial velocity into horizontal and vertical components. This can be done using basic trigonometry.

We take the initial velocity to be $v$, and make this the hypotenuse of our triangle. The launch angle of the projectile is our angle, $\theta$.

Note: Diagram not to scale.

For this problem, we require only the vertical velocity component.

As we know that $\sin \left(\theta\right) = \frac{y}{r}$ (where $r$ is the hypotenuse), we can solve for ${v}_{y}$ as: ${v}_{y} = r \sin \left(\theta\right)$.

${v}_{y} = r \sin \left(\theta\right)$

${v}_{y} = 37 \cdot \sin \left(\frac{\pi}{12}\right)$

${v}_{y} = 9.6 \frac{m}{s}$

We will assume this to be the initial vertical velocity of the projectile, while the final vertical velocity of the projectile will be $0$ at its maximum altitude (at this point it will stop momentarily before falling back to Earth).

Because this object is in free-fall, its vertical acceleration is equal to $- g$, the free-fall acceleration, where $g = 9.8 \frac{m}{s} ^ 2$. Thus, we can use the kinematic equation above to find the maximum height of the projectile.

(v_f)_y^2=(v_i)_y^2+2a_yΔy

Rearrange to solve for Δy:

Δy=((v_f)_y^2-(v_i)_y^2)/(2a_y)

Δy=-(9.6m/s^2)^2/(2(-9.8m/s^2))

Δy=4.7m

The maximum altitude of the projectile is $4.7 m$.