# A projectile is shot at a velocity of  2 m/s and an angle of pi/4 . What is the projectile's peak height?

Feb 5, 2017

${y}_{\text{max"(0.14)~~0.10" meters}}$ (that's about 4 inches)

#### Explanation:

You have a projectile and you want to know the vertical position (i.e., the $y$-value) at the projectile's peak. The general equation is

$y = \left({v}_{0} \sin \left({\theta}_{0}\right)\right) t - \frac{1}{2} g {t}^{2}$

You are given

• ${v}_{0} = 2$ m/s (pretty slow, about 4.5 mph)
• ${\theta}_{0} = \pi \text{/} 4$
• $g = 9.8$ m/s""^2

Plugging in those values we get $y \left(t\right)$

$y \left(t\right) = \left(2 {\text{ m/s")sin(pi"/"4)t-0.5(9.8" m/s}}^{2}\right) {t}^{2}$
$\text{ } = \left(\sqrt{2}\right) t - 4.9 {t}^{2}$ meters

A graphing calculator will find your max height at $y \approx 0.10$
graph{y=2sin(pi/4)x-0.5(9.8)x^2[-.01,.3,-.01,.11]}

The time at which the this function is maximum is discovered by solving for $t$ when $\mathrm{dy} \text{/} \mathrm{dt} = 0$.

$\frac{d}{\mathrm{dt}} \left(y\right) = y ' \left(t\right) = \sqrt{2} - 9.8 t$

Set $y ' \left(t\right) = 0$

$\sqrt{2} - 9.8 t = 0$

Now solve for $t$
$9.8 t = \sqrt{2}$
$t = \sqrt{2} \text{/} 9.8$
$t \approx 0.14$ seconds

So the maximum height occurs at that time

y_"max"(0.14)~~(2"m/s")sin(pi"/"4)(0.14"s")-1"/"2(9.8"m/s")(0.14"s")^2
${y}_{\text{max"(0.14)~~0.10" meters}}$ (that's about 4 inches)