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)
Answers
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.
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.