# If a projectile is shot at an angle of pi/4 and at a velocity of 15 m/s, when will it reach its maximum height??

Feb 28, 2016

2.12s

#### Explanation:

Velocity of projection$, v = 15 m {s}^{-} 1$
Angle of projection$, \alpha = \frac{\pi}{4}$
Vertical component of vel. of projection$, v \sin \alpha = 15 \cdot \sin \left(\frac{\pi}{4}\right) = \frac{15}{\sqrt{2}} m {s}^{-} 1$
Applying equn for vertical motion under gravity,
at maximum height (H) the final vertical vel. being ZERO, we can write
${0}^{2} = {\left(v \sin \alpha\right)}^{2} - 2 \cdot g \cdot H$
$\implies H = {\left(v \sin \alpha\right)}^{2} / \left(2 g\right) = {15}^{2} / \left({2}^{2} \cdot 10\right) = 5.625 m$, [taking g = 10$m {s}^{-} 2$]
If time of reaching Maximum height be T then
$0 = v \sin \alpha \cdot T - \frac{1}{2} g {T}^{2}$
$\implies T = \frac{2 v \sin \alpha}{g} = 2 \cdot \frac{15}{10 \sqrt{2}} = 2.12 s$

Calculation of height not wanted here