Question

A tennis player tosses a tennis ball straight up and then catches it after 1.57 s at the same height as the point of release.

Answer 1

PART A:

-9.8 m/s^2 (S)

PART B:

0 meters (N)

PART C:

≈ 7.7 (ROUNDED)
Accurate: 7.69300

PART D:

4.9 m (Δy)

Transcribed Text:
A tennis player tosses a tennis ball straight up and then catches it after 1.57 seconds at the same height as the point of release.

Required:

• Acceleration of the ball while in flight
• Velocity of the ball at maximum height
• Initial velocity
• Maximum height reached

Use the following kinematic equations:

`v=v_o+at`

Final velocity = Initial Velocity + Acceleration × Time

Δ
`x=v_ot+(1)/(2)at^2`

Change in position (displacement) = Initial velocity × time + (0.5) (acceleration × final time)²

`v^2=v_o2+2a`Δ
`x`

Final velocity (squared) = Initial velocity (two times) + (two times) acceleration * change in position

*A illustration shall be provided for better understanding*

Part A:

Once the tennis player releases the ball, the only force acting on that ball is the gravitational force which exerts an acceleration of -9.8 m/s^2.

Part B:

At that maximum height (as seen in the attached picture) would have a velocity of 0 for just a split moment making this the answer.

PART C: *Be aware that the PART A and B are just conceptual type questions and now we have actual calculations to figure out*.

Reviewing what we know,

Acceleration = -9.8 m/s^2

Time = 2.00 seconds

Now we have to find the initial velocity.

Reviewing the picture and the final velocity, it should turn out this way:

• 
  `Vi=V_i,V=-V_o`

Looking at the kinematic equations we would use:

`V_f=V_i+at`

Current equation:

`-V_o=V_o+(-9.8)(1.57)`

* MULTIPLY -9.8 and 1.57!!!

Now we must subtract the V(naught) from both sides.

The left and right side would become:

`-2v_o=-15.38600`

Divide -2 from both sides.

Initial velocity = 7.69300

Rounded

≈ 7.7 m/s

This would make the initial velocity equal to 7.7 m/s

PART D:
In order to solve for maximum height, we must use some values we would need to find this answer.

Initial velocity = 7.69300 m/s

Acceleration = -9.8m/s^2

Final velocity (or velocity at max height) = 0 m/s

Δy = ...

The kinematic equation we would be using is:

`v^2=v_o2+2a`Δ
`x`

Because the situation is in the vertical dimension, the 'delta x' will be changed to:

`v^2=v_o2+2a`Δ
`y`

Subtract the initial velocity squared from both sides of the equation.

This would give us:

`v^2-v_0^2=2a`Δ
`y`

Divide both sides by 2a.

By finding this we can find the vertical displacement.

`(v^2-u^2)/(2a)`=Δ
`y`

Let plug in the known values

`((0)^2-(9.8)^2)/(2(-9.8^2)) = 4.9m`

Done