Question

A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall downwards (take g=9.8m/s square)

Answer 1

By using equation of motion
S = u² / (2g) ………[∵ Velocity at maximum height is 0]
= (15 m/s)² / (2 × 9.8 m/s²)
= 11.48 m

It will travel height of 11.48 m before it begins to fall.

Answer 2

Given: The speed with which the ball is thrown (u) = 15 m/s

g = 9.8 m/s²

To Find: Height it attains before it begins to go downwards

Solution:

Let us assume the direction towards the ground as negative and direction towards the sky as positive

The acceleration acting on the ball is acceleration due to gravity(g)

g acts in downward direction at all times

When the ball is thrown upwards its speed is opposite to g, so its speed decreases continuosly till it reaches 0 after that the speed becomes negative

When the speed is zero then the body is at the highest point

Using the equation of motion

v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance travelled

(0)² = (15)² + 2 (-9.8)(h)

225 = 19.6 h

h = 11.48 m

Therefore, the height attained by the body is 11.48m.