Projectile Motion with only the vertical/ horizontal height and angle given

  • Thread starter Otakuchica
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  • #1

Main Question or Discussion Point

I need help! I understand regular projectile, but when it comes to using an angle it drives me crazy!
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
 

Answers and Replies

  • #2
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Which quantities are given and which did you find out (and how)?
It would be nice to show the problem you are trying to solve.
 
  • #3
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1
I need help! I understand regular projectile, but when it comes to using an angle it drives me crazy!
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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