A tennis ball is dropped from the top of a building. It falls for 2.5 seconds before it hits the ground. (a) What velocity is the tennis ball traveling right before it hits the ground? (b) How high is the building?

Try using the formulas

V = gt and
Y = (1/2) g t^2

Hi

To answer these questions, we need to use the equations of motion for a falling object under the influence of gravity.

(a) What velocity is the tennis ball traveling right before it hits the ground?

The equation we can use to determine the velocity is:

v = u + gt

Where:
- v is the final velocity
- u is the initial velocity (0 m/s in this case as the ball was dropped)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken (2.5 seconds in this scenario)

Using the equation, we can calculate the final velocity:

v = 0 + (9.8 m/s^2) * (2.5 s)
v = 0 + 24.5 m/s
v = 24.5 m/s

Therefore, the tennis ball is traveling at a velocity of 24.5 m/s before hitting the ground.

(b) How high is the building?

To calculate the height of the building, we can use the equation:

h = ut + (1/2)gt^2

Where:
- h is the height of the building
- u is the initial velocity (0 m/s)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time taken (2.5 seconds)

Substituting the given values, we can find the height:

h = 0 * (2.5 s) + (0.5) * (9.8 m/s^2) * (2.5 s)^2
h = 0 + 0.5 * 9.8 * 6.25
h = 0 + 4.9 * 6.25
h = 0 + 30.625
h = 30.625 m

Therefore, the height of the building is 30.625 meters.

To answer these questions, we need to use the equations of motion for a falling object. There are three key equations we can use:

1. Displacement equation: Δy = v₀t + (1/2)at²
2. Final velocity equation: v = v₀ + at
3. Average velocity equation: v_avg = (v₀ + v) / 2

Here's how we can use these equations to find the answers:

(a) What velocity is the tennis ball traveling right before it hits the ground?

To find the velocity right before the tennis ball hits the ground, we can use the second equation. We know that the initial velocity is 0 (since the ball was dropped) and the time is 2.5 seconds. The acceleration due to gravity (a) is approximately 9.8 m/s².

Plugging in the values:

v = v₀ + at
v = 0 + 9.8 * 2.5
v = 24.5 m/s

Therefore, the ball is traveling at a velocity of 24.5 m/s right before it hits the ground.

(b) How high is the building?

To find the height of the building, we can use the first equation. Again, the initial velocity is 0 and the time taken is 2.5 seconds.

Plugging in the values:

Δy = v₀t + (1/2)at²
Δy = 0 * 2.5 + (1/2) * 9.8 * (2.5)²
Δy = 0 + (1/2) * 9.8 * 6.25
Δy = 3.125 * 9.8
Δy = 30.625 m

Therefore, the height of the building is 30.625 meters.