- #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

- #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

- #2

nasu

Gold Member

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- 433

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

- #3

- 43

- 1

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.

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