# A ball's thrown upwards from the roof of a building with a speed of 15 m/s. Velocity of the ball when it's 5 m above the roof?

Feb 20, 2017

$\textsf{+ 11.27 \textcolor{w h i t e}{x} \text{m/s}}$ and $\textsf{- 11.27 \textcolor{w h i t e}{x} \text{m/s}}$

#### Explanation:

We can use:

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

This becomes:

$\textsf{s = u t - \frac{1}{2} \text{g} {t}^{2}}$

$\therefore$$\textsf{5 = 15 t - \frac{1}{2} \times 9.81 {t}^{2}}$

$\therefore$$\textsf{4.905 {t}^{2} - 15 t + 5 = 0}$

$\textsf{t = \frac{15 \pm \sqrt{225 - 4 \times 4.905 \times 5}}{9.81}}$

This gives 2 values for $\textsf{t}$:

$\textsf{{t}_{1} = 2.677 \textcolor{w h i t e}{x} s}$

and

$\textsf{{t}_{2} = 0.38 \textcolor{w h i t e}{x} s}$

These 2 times refer to the object moving up and then returning back down.

Using the convention "up is +ve" we can use:

$\textsf{v = u + a t}$

This becomes:

$\textsf{v = u - \text{g} t}$

Using $\textsf{{t}_{2} \Rightarrow}$

$\textsf{v = 15 - 9.81 \times 0.38 = + 11.27 \textcolor{w h i t e}{x} \text{m/s}}$

This is the velocity on the way up.

Using $\textsf{{t}_{2} \Rightarrow}$

$\textsf{v = 15 - 9.81 \times 2.677 = - 11.27 \textcolor{w h i t e}{x} \text{m/s}}$

This is the velocity on the way down.

What goes up, must come down.