Question #60fb7

1 Answer
Jan 14, 2018

1:1

Explanation:

enter image source here With in 1st 3 sec,kinetic energy will be (1/2)m(v^2) or (1/2)m(u-g*t)^2 = KE1 (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*t or (10*6) or 60 m/s

So,KE1 = (1/2)m(60-(10*3))^2 or 900(1/2)m

At the highest point it's potential energy is mgh or (mg*(u^2)/(2g)) or (1/2)m(60)^2 or 3600(1/2)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 mgh1 or mg(((u^2)-(v^2))/(2g)) or 2700(1/2)m( where, h1 is the height reached in 3 sec)

So,with in these 3 sec increase in potential energy=decrease in kinetic energy=KE2=(3600-2700)(1/2)m or 900(1/2)m
So, KE1:KE2 = 1:1