# Question #afdf8

Jun 23, 2015

The answer is b) 65 m/s.

#### Explanation:

Here's how you should approach this problem.

You know that your ball is projected upwards at an initial velocity which we'll call ${v}_{0}$.

You also know that the distance the ball covered in its ${5}^{t h}$ second of flight is twice the distance it covered in its ${6}^{t h}$ second of flight.

Since the ball is projected upwards, this difference between the distances makes sense because the ball is moving against gravity, which implies that it's slowing down.

To get the distance the ball travelled in its ${5}^{t h}$ second, calculate the distance it covered after 4 seconds of flight and subtract this distance from the distance it covered after 5 seconds of flight.

SIDE NOTE I'll skip adding the units in order to keep the text clean.

${d}_{\text{4 sec}} = {v}_{0} \cdot 4 - \frac{1}{2} \cdot 10 \cdot {4}^{2} = 4 {v}_{0} - 80$

${d}_{\text{5 sec}} = {v}_{0} \cdot 5 - \frac{1}{2} \cdot 10 \cdot {5}^{2} = 5 {v}_{0} - 125$

The distance it covered in its ${5}^{t h}$ second was

$\Delta {d}_{4 , 5} = {d}_{\text{5 sec" - d_"4 sec}}$

$\Delta d \left(4 , 5\right) = 5 {v}_{0} - 125 - 4 {v}_{0} + 80 = {v}_{0} - 45$

Do the same to get the distance covered in its ${6}^{t h}$ second

${d}_{\text{6 sec}} = {v}_{0} \cdot 6 - \frac{1}{2} \cdot 10 \cdot {6}^{2} = 6 {v}_{0} - 180$

${d}_{\text{5 sec}} = 5 {v}_{0} - 125$

Therefore,

$\Delta {d}_{5 , 6} = {d}_{\text{6 sec" - d_"5 sec}}$

$\Delta {d}_{5 , 6} = 6 {v}_{0} - 180 - 5 {v}_{0} + 125 = {v}_{0} - 55$

Now use the fact that

$\Delta {d}_{4 , 5} = 2 \cdot \Delta {d}_{5 , 6}$

${v}_{0} - 45 = 2 \cdot \left({v}_{0} - 55\right)$

${v}_{0} - 45 = 2 {v}_{0} - 110$

${v}_{0} = 110 - 45 = \textcolor{g r e e n}{\text{65 m/s}}$