Question

A ball is projected vertically upwards with an initial velocity 'u goes to a maximum height 'h before touching the ground. What is the value of 'h?

Answer 1

By using formula
v^2=u^2+2as
we can find the answer.

here, v=0, s=h, a=g.

.: u^2+2gh=0
.: u^2=-2gh
.: h=u^2/g.

Don't get confused with -ve sign. it indicates deceleration.

Answer 2

Given:

• The initial velocity of the ball = 'u
• Maximum height of the ball before touching the ground = 'h

To Find:

• The value of 'h.

Solution:

There are three formulas for the projectile motion of an object in physics. Considering one of the formulas to find the value of 'h.

⇒$v^2=u^2+(2*(-g)*h)$   → (equation 1)

Where 'v' is the final velocity, 'u' is the initial velocity, 'g' is the acceleration due to gravity, and 'h' is the height.

Note: In equation 1, acceleration due to gravity is negative because the ball is projected in the upwards direction which is opposite to the direction of acceleration due to gravity and this decreases the final velocity to 0 m/s.

Now re-arranging equation 1 we get

⇒ $v^2=u^2-2gh$

⇒ $v^2-u^2 = 2gh$

⇒ $h = 'h = \frac{v^2-u^2}{2g}$

∴ The value of 'h = $\frac{v^2-u^2}{2g}$