If a 7kg object is thrown straight upwards at a speed of 10m/s(meters per second)

To what height will it rise? note: gravitational constant = -9.8m/s

(In reply to Puzzle Solution by K Sengupta)

Let us denote the kinetic energy of the ball in its course of the motion before coming to rest as K(E). We will represent the potential energy of the ball after completing its upward journey as P(E).

Then, K(E) =  (1/2)*m*v^2, and: P(E) = m*g*h, where:
g = absolute magnitude of the gravitational constant = 9.8 m/s
v = initial velocity of the ball
h = the height to which the ball will rise.

In terms of the laws of conservation of energy, we must have:

K(E) = P(E)
-> (1/2)*m*v^2 = m*g*h
-> h = (v^2)/(2*g) = 100/(2*9.8) ~ 5.1

Consequently, the ball will rise to a height of approximately 5.1 meters .

Posted by K Sengupta on 2008-12-04 06:17:37