Question

A ball is thrown upward with a speed of 28.2 m/s.

A. What is its maximum height?

B. How long is the ball in the air?

C. When does the ball have a speed of 12m/s? (there are two answers, give both)

Answer 1

A.

Initial velocity of the ball is, .

Let be the maximum height reached by the ball.

Acceleration due to gravity,   (since it is pointing downward)

Velocity of the ball will become zero when the ball reaches its maximum height .

From equation of motion,

, where   and   are initial and final velocity of the particle, and is the uniform acceleration.

Since   becomes zero at height h, we have,

Therefore maximum height reached by the ball is

B.

From equation of motion we have,

where and   are initial and final position of the ball.

Since we are calculating how long is the ball in the air, which is the time taken by the ball from leaving the ground to reaching the ground. Hence

Hence ball will be in the air for

C.

Consider, final velocity, . There arises two situation.

1) From height 0 to h.(i.e. when the ball is moving upwards)

From equation of motion,

here , and initial velocity,

2) From height h to 0.(i.e. during falling)

First let us calculate the time required to reach the maximum height .

Here , and initial velocity, .

(The same time will be required by the ball to reach the ground, hence twice of this time would give us back the time of flight)

Now   (negative because velocity of the ball is in the downward direction), and since ball is falling from maximum height, its initial velocity, will be zero.

The total time required by the ball to reach velocity, while falling is the sum of the time taken by the ball to reach maximum height and the time taken to reach velocity, .

Hence total time required to reach velocity, is,

Hence the ball will have velocity 12m/s at time, and  .