# A projectile is shot from the ground at an angle of pi/4  and a speed of 7 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Jan 16, 2017

The answer is $= 2.5 m$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

$u = 7 \sin \left(\frac{\pi}{4}\right) m {s}^{-} 1$

$v = 0$

$a = - 9.8 {m}^{- 1}$

We use the equation

$v = u + a t$

$0 = 7 \sin \left(\frac{\pi}{4}\right) - 9.8 \cdot t$

$t = \frac{7}{9.8} \cdot \sin \left(\frac{\pi}{4}\right) = 0.505 s$

Resolving in the ${\rightarrow}^{+}$

$u = 7 \cos \left(\frac{\pi}{4}\right) m {s}^{-} 1$

$x = 7 \cos \left(\frac{\pi}{4}\right) \cdot 0.505 = 2.5 m$