# Question #60fb7

Jan 14, 2018

1:1

#### Explanation: With in 1st 3 sec,kinetic energy will be $\left(\frac{1}{2}\right) m \left({v}^{2}\right)$ or $\left(\frac{1}{2}\right) m {\left(u - g \cdot t\right)}^{2}$ = $K E 1$ (u is the initial velocity)

Now,given within 6 sec it reaches the highest point,so initial velocity with which it was thrown was $g \cdot t$ or $\left(10 \cdot 6\right)$ or 60 m/s

So,KE1 = $\left(\frac{1}{2}\right) m {\left(60 - \left(10 \cdot 3\right)\right)}^{2}$ or $900 \left(\frac{1}{2}\right) m$

At the highest point it's potential energy is $m g h$ or $\left(m g \cdot \frac{{u}^{2}}{2 g}\right)$ or $\left(\frac{1}{2}\right) m {\left(60\right)}^{2}$ or $3600 \left(\frac{1}{2}\right) m$ ; where, $h$ is the maximum height reached by the ball. (i.e by the next 3 sec its kinetic energy got totally converted to potential energy)

Now,after 3 sec its potential energy was $m g h 1$ or $m g \left(\frac{\left({u}^{2}\right) - \left({v}^{2}\right)}{2 g}\right)$ or $2700 \left(\frac{1}{2}\right) m$( where, $h 1$ is the height reached in 3 sec)

So,with in these 3 sec increase in potential energy=decrease in kinetic energy=$K E 2$=$\left(3600 - 2700\right) \left(\frac{1}{2}\right) m$ or $900 \left(\frac{1}{2}\right) m$
So, $K E 1 : K E 2 = 1 : 1$