Question

A man throws ball with a velocity of 20 m/s upwards . The ball decelarates with 10m/s². Find the maximum height reached by ball.​

Answer 1

Answer:

20 m.

Explanation:

In the given question,

Initial velocity(u) = 20 m/s

acceleration(a) = - deceleration = - 10 m/s²

At the highest point/maximum height, under the influence of gravity velocity the ball will be zero.

final velocity(v) = 0

Using equation of motion:

=> v² = u² + 2aS

=> 0² = 20² + 2(-10)S

=> 20S = 400

=> S = 400/20

=> S = 20

Maximum height reached by the ball is 20 m.

Answer 2

Given :-

A man throws ball with a velocity of 20 m/s upwards . The ball decelerates with 10m/s².

To Find :-

Maximum height

Solution :-

We know that

v² = u² + 2as

0² = 20² + 2(-10)s

0 = 400 + (-20s)

-400 = -20s

400 = 20s

400/20 = s

20 = s