A tennis ball is thrown horizontally from an elevation of 20.60 m above the ground with a speed of 21.0 m/s?

Where is the ball after 2.00 s?

vertical m from the ground?
If the ball is still in the air, how long before it hits the ground? (If the ball is not in the air, enter 0.)
s?
If the ball is still in the air, where will it be with respect to the starting point once it lands? m

please helpp thnks

horizontal speed = 21 until it hits ground

so
x = 21 t as long as it is airborne

v = - g t = -9.8 t
y = 20.6 - 4.9 t^2

You should be able to get your answers from that.

To answer these questions, we can use the equations of motion for projectile motion. In this case, since the tennis ball is thrown horizontally, there is no initial vertical velocity.

1. Where is the ball after 2.00 s?

To find the horizontal distance traveled by the ball after 2.00 seconds, we can use the formula:

horizontal distance = initial horizontal velocity * time

Given that the initial horizontal velocity is 21.0 m/s and the time is 2.00 seconds, we can calculate the horizontal distance:

horizontal distance = 21.0 m/s * 2.00 s = 42.0 m

Therefore, the ball is 42.0 meters away horizontally from the starting point after 2.00 seconds.

2. Vertical distance from the ground?

Since the ball is thrown horizontally, there is no vertical acceleration. We can use the formula:

vertical distance = initial vertical velocity * time + (1/2) * acceleration * time^2

In this case, the initial vertical velocity is 0 m/s, the time is 2.00 seconds, and the acceleration due to gravity is approximately 9.8 m/s^2.

vertical distance = 0 m/s * 2.00 s + (1/2) * 9.8 m/s^2 * (2.00 s)^2
vertical distance = (1/2) * 9.8 m/s^2 * 4.00 s^2 = 19.6 m

Therefore, the ball is 19.6 meters above the ground after 2.00 seconds.

3. Time before it hits the ground?

To find the time before the ball hits the ground, we can use the equation:

vertical distance = (1/2) * acceleration * time^2

In this case, the vertical distance is 20.60 m (initial elevation), the acceleration is -9.8 m/s^2 (negative because it's downward), and we need to find the time.

Plugging in the values, we get:

20.60 m = (1/2) * (-9.8 m/s^2) * time^2

Simplifying the equation, we have:

10.30 m = -4.9 m/s^2 * time^2

Dividing both sides by -4.9 m/s^2, we get:

time^2 = -10.30 m / -4.9 m/s^2 = 2.10 s^2

Taking the square root of both sides gives us:

time = √(2.10 s^2) ≈ 1.45 s

Therefore, it would take approximately 1.45 seconds before the ball hits the ground.

4. Where will the ball be with respect to the starting point once it lands?

Since the ball was thrown horizontally, it will continue moving horizontally until it lands. Therefore, the horizontal distance from the starting point will remain the same.

The ball travels a horizontal distance of 42.0 meters after 2.00 seconds, so once it lands, it will be 42.0 meters horizontally away from the starting point.