# A ball is projected with a horizontal velocity of 5.1 ms^-1 at a height of 2.5 xx 10^2 m. How far does it travel before it hits the ground?

Jun 24, 2017

The horizontal distance traveled will be $s = {u}_{x} t = 5.1 \times 7.1 = 36.2$ $m$.

#### Explanation:

As for all projectile motion questions, we can treat the horizontal and vertical direction independently. If we ignore air resistance, the horizontal velocity will remain constant at ${u}_{x} = 1.5$ $m {s}^{-} 1$.

We need to calculate the time taken for the ball to fall. The initial velocity ${u}_{y}$ in the vertical direction is $0$ $m {s}^{-} 1$.

$s = {u}_{y} t + \frac{1}{2} {a}_{y} {t}^{2}$

$t = \sqrt{\frac{2 s}{a}}$ (since ${u}_{y} = 0$ and we can rearrange to make $t$ the subject)

$t = \sqrt{\frac{2 \left(2.5 \times {10}^{2}\right)}{9.8}} = 7.1$ $s$

The horizontal distance traveled in this time will be $s = {u}_{x} t = 5.1 \times 7.1 = 36.2$ $m$.