Finding Max Height and Velocity of Thrown Kong

• Charlie_russo
In summary, for the physics experiment involving throwing a kong as far as possible, the instructions were to record the distance and time of two throws. Using the formula V=d/t, the horizontal velocities for both throws were calculated. To find the vertical velocity, the maximum height of the kong needs to be determined first. This can be found by using the formula h=V^2/2g, where V is the initial velocity and g is the acceleration due to gravity. Once the maximum height is known, the vertical velocity can be calculated by dividing the maximum height by half of the time. For the first throw, the horizontal and vertical velocities were 10.61m/s and 12.25m/s, respectively, and
Charlie_russo
For a physics experiment, we had to take a kong (dog toy with a rubber ball on the end of a rope), twirl it and then throw it as far as we could. We were told to record how long it was in the air for, and the distance for two throws. This was all was we were told to record. Afterwards, we need to find the maximum height the kong reaches, it's horizontal velocity as it is in the air and it's vertical velocity when it hits the ground.

Throw #1-

Distance: 52m
Time: 4.9s

Throw #2-

Distance: 65m
Time: 3.84s

So far, I have been able to figure out the horizontal velocity by using V=d/t.

#1- 52m/4.9s=10.61m/s

#2- 65m/3.84s=16.93m/s

Now, I think that in order to find the vertical velocity, I would need to first find the kongs maximum height, and then divide this by half of the time, (since at the halfway point the kong will now be traveling horizontally, and would take the same amount of time to reach the ground as a kong that is dropped from the same height). However, I am not sure on how to find the maximum height.

ok, here is what i got...

#1-

Horizontal Velocity- 10.61m/s
Vertical Velocity- 12.25m/s
Maximum Height- 30.01m

#2-

Horizontal Velocity- 16.93m/s
Vertical Velocity- 9.6m/s
Maximum Height- 18.43m

Hello,

Thank you for sharing your experiment and data. It sounds like you have made good progress in finding the horizontal velocity of the thrown kong. To find the maximum height and vertical velocity, we will need to use some basic equations from kinematics.

First, we can use the equation d = v0t + (1/2)at^2 to find the vertical distance traveled by the kong. In this equation, d is the distance, v0 is the initial velocity (which is 0 since the kong starts at rest), a is the acceleration (which is due to gravity and has a value of -9.8 m/s^2), and t is the time.

For Throw #1, using the given time of 4.9s, we can calculate the vertical distance as follows:

d = (0)(4.9) + (1/2)(-9.8)(4.9)^2 = -120.05m

Since the kong reaches its maximum height at the halfway point (2.45 seconds), we can divide this distance by 2 to get the maximum height:

Maximum height = -120.05m/2 = 60.03m

We can use the same process for Throw #2 to find the maximum height of the kong.

Now, to find the vertical velocity when the kong hits the ground, we can use the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. Since we know that the final velocity when the kong hits the ground is 0 m/s, we can rearrange the equation to solve for vi:

vi = -at

Using the time for Throw #1 (4.9s) and the acceleration due to gravity (-9.8 m/s^2), we can calculate the initial vertical velocity as:

vi = -(9.8)(4.9) = -48.02m/s

We can use the same process for Throw #2 to find the initial vertical velocity.

I hope this helps you in finding the maximum height and vertical velocity of the thrown kong in your experiment. Keep up the great work in your physics experiments!

What is the formula for finding the maximum height of a thrown object?

The formula for finding the maximum height of a thrown object is h = (v02sin2θ)/2g, where v0 is the initial velocity, θ is the angle of the throw, and g is the acceleration due to gravity.

How do you calculate the maximum velocity of a thrown object?

The maximum velocity of a thrown object can be calculated using the formula vmax = v0sinθ, where v0 is the initial velocity and θ is the angle of the throw.

What factors can affect the maximum height and velocity of a thrown object?

The maximum height and velocity of a thrown object can be affected by factors such as air resistance, wind, and the angle and force of the throw.

How is the maximum height and velocity of a thrown object related?

The maximum height and velocity of a thrown object are directly related. This means that as the maximum height increases, the maximum velocity also increases. This relationship is described by the equation vmax = √(2gh), where h is the maximum height and g is the acceleration due to gravity.

Can the maximum height and velocity of a thrown object be calculated for any object?

Yes, the maximum height and velocity of a thrown object can be calculated for any object as long as the initial velocity and angle of the throw are known. However, the calculations may not be accurate for objects with irregular shapes or significant air resistance.

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