Question

A ball is thrown up with a speed of 5 m/sec. How high will it go before falling down.

Answer 1

u = 5 m/s
g = -10 m/s^2 (downward direction is negative)
v = 0 (at highest point)
H= ?

We know that v^2 - u^2 = 2gH

0^2 - 5^2 = 2×(-10)×H

-25 = -20 H

H = 25/20 = 1.25 m

So the ball will go 1.25m up before falling down.

Answer 2

By Newton's third kinematical equation

v²=u²+2gs

plug in v=0, u=15, g=-9.8(negative since the ball is decelerating)

You get s=11.47959183673469

Method 2

When it is being thrown assume it has 0 potential energy, and maximum Kinetic energy. Since the kinetic energy is being converted into potential energy, we can say

K.E=P.E

½mv²=mgh

v=√(2gh) and h=11.47959183673469 in this method u have to do this plz mark my answer as brainlist