Question

A boy throws a ball with a velocity of 30m/s.Find the total tome elapsed between throwing and catching the ball

(take g=10m/s2)

Answer 1

By applying the law of conservation of energy, we can write the following equation.

Gain in potential energy =loss in kinetic energy.

mgh = 1/2(mu^2)

Where, the symbols represent usual quantities.

Hence, h = u^2/2g = 30^2/2x10 = 900/20

= 45 m

Hence, the maximum height attained is 45 m.

Alternative method:

v^2 = u^2 +2gh

0 = 30^2 +2(-10)h, at the highest point v=0

20h = 900

h = 900/20 = 45 m

Answer 2

x=45.92m

Step-by-step explanation:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

With these questions it's best to just memorise the suvat equations and either draw or imagine the actions involved, that was you can tell what piece of given information translates to which variable. For example; I know that I am looking for max height, so when a ball is at its highest point, it can't be moving up anymore (thus V = 0). I also know that it is moving upwards against gravity, so gravity will be decelerating the ball (therefore a=-g). In a paper, it may as for assumptions with this question, a good answer would be no air resistance or movement due to any other external forces. I hope I helped :)