- #1

- 1

- 0

The vertical distance is 15.01 m

Horizontal distance is 21.84 m

Angle is 45°

I figured the time to be 2.1 sec

I need the velocity

And V

*sub*F is 0...right?

I just need the help guys, thanks

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- Thread starter Otakuchica
- Start date

In summary, the person is struggling with using an angle in projectile motion and needs help finding the velocity. They have determined the vertical and horizontal distance and time, and are unsure if VsubF is 0. They are advised to break the trajectory in half and check their calculated time. The conversation also mentions finding the initial velocity using Pythagoras' theorem. The person is asked to provide more details about the problem they are trying to solve.

- #1

- 1

- 0

The vertical distance is 15.01 m

Horizontal distance is 21.84 m

Angle is 45°

I figured the time to be 2.1 sec

I need the velocity

And V

I just need the help guys, thanks

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- #2

Homework Helper

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It would be nice to show the problem you are trying to solve.

- #3

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Otakuchica said:

The vertical distance is 15.01 m

Horizontal distance is 21.84 m

Angle is 45°

I figured the time to be 2.1 sec

I need the velocity

And VsubF is 0...right?

I just need the help guys, thanks

Don't worry about the angle. V and H motion are still independent. Break the trajectory in half, and let's check if your time is correct. A falling object falls as d=0.5gt^2, this one fell 15m, if G≈10m/s^2 then this one fell:

15m=0.5*9.8(m/s^2)*t^2 --> t=√15/4.9=1.7496s

Vhorizontal=Δx/t=21,84m/(2*1.7496s)=6.241m/s (the time is doubled because the time in the first step is only for the falling part of the motion.

Vinitialvertical=Vhorizontal

Vinitialtotal=Vhorizontal*√2 by Pythagoras.

If those aren't what you're looking for, you need to describe the problem more.

Projectile motion refers to the motion of an object that is projected into the air and then moves under the influence of gravity. It is a combination of horizontal and vertical motion.

The angle of projection is directly related to the range of a projectile. The range will be maximum when the angle of projection is 45 degrees.

Yes, the time of flight can be calculated using the formula t = (2 * vertical height / g), where g is the acceleration due to gravity (9.8 m/s²).

The final velocity of a projectile can be found using the formula v = sqrt(2 * g * vertical height), where g is the acceleration due to gravity (9.8 m/s²).

Yes, the maximum height of a projectile can be found using the formula h = (vertical height)^2 / (2 * g), where g is the acceleration due to gravity (9.8 m/s²).

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