Suppose that an object is dropped from a height of h meters and hits the ground with a velocity of v meters per second. Then v= sqrt19.6h. If an object hits the ground with a velocity of 24.2 meters per second, from what height was it dropped?

Question

Illustrate an object being dropped from a height, represented in a visually abstract manner. The object is flying through the air and about to hit the ground. Behind it, a fast fading trail, showing the trajectory of the fall. The backdrop of the image should emphasize the height of the drop. Also, incorporate visual emphasis on the speed of the object as it descends, as if it is travelling at 24.2 meters per second. Don't include any text or numbers in the image, and remember to give the picture an overall sense of depth and motion.

Suppose that an object is dropped from a height of h meters and hits the ground with a velocity of v meters per second. Then v= sqrt19.6h. If an object hits the ground with a velocity of 24.2 meters per second, from what height was it dropped?

Carry your intermediate computations to at least four decimal places, and round your answer to the nearest tenth.

Answer 1

If your rounding to the nearent tenth though that can't plug in. I'm so confused.

Answer 2

just plug in your value:

24.2 = √(19.6 h)
√h = 24.2/√19.6
h = (24.2/√19.6)^2 = 29.88 m

or

I know from physics that
h = 4.905 t^2 gives you the height from which an object was dropped after t seconds ,
and v = 9.81 t gives you the velocity after t seconds

so 24.2 = 9.81 t
t = 2.46687..
h = 4.905(2.46687..)^2= appr 29.85 m

Answer 3

Suppose a pendulum is L meters long. The time, t, in seconds that it takes to swing back and forth once is given by t=2.01\sqrt()L. If a pendulum is 20.25 meters long, how long does it take to swing back and forth once?

Round your answer to the nearest tenth.

Answer 4

Why did the object fall on the ground with such velocity? Because it couldn't resist the pull of gravity!

Now, let's get back to the problem. We know that the velocity is given by v = sqrt(19.6h). If the object hits the ground with a velocity of 24.2 m/s, we can set up the equation:

24.2 = sqrt(19.6h)

To find the height, we need to isolate h. So let's square both sides of the equation:

(24.2)^2 = (sqrt(19.6h))^2

586.44 = 19.6h

Divide both sides by 19.6:

h = 586.44/19.6

h ≈ 29.9

So, the object was dropped from a height of approximately 29.9 meters. I hope this answer didn't drop the ball on humor!

Answer 5

To find the height from which the object was dropped, we can use the equation v = √(19.6h), where v is the velocity in meters per second and h is the height in meters.

Given that the velocity of the object is v = 24.2 meters per second, we can substitute this value into the equation:

24.2 = √(19.6h)

To solve for h, we can square both sides of the equation:

(24.2)^2 = (√(19.6h))^2

585.64 = 19.6h

Now, divide both sides of the equation by 19.6 to isolate h:

h = 585.64 / 19.6

Calculating this, we get:

h ≈ 29.8796 meters

Rounding this to the nearest tenth, the object was dropped from a height of approximately 29.9 meters.

Answer 6

Suppose that an object is dropped from a height of h meters and hits the ground with a velocity of v meters per second. Then v= sqrt19.6h. If an object hits the ground with a velocity of 24.2 meters per second, from what height was it dropped?

If your rounding to the nearent tenth though that can't plug in. I'm so confused.

just plug in your value:

Suppose a pendulum is L meters long. The time, t, in seconds that it takes to swing back and forth once is given by t=2.01\sqrt()L. If a pendulum is 20.25 meters long, how long does it take to swing back and forth once?

Why did the object fall on the ground with such velocity? Because it couldn't resist the pull of gravity!

To find the height from which the object was dropped, we can use the equation v = √(19.6h), where v is the velocity in meters per second and h is the height in meters.

The answer is 32.1 meters