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

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SUMMARY

The discussion centers on calculating the velocity of a projectile given its vertical distance of 15.01 m, horizontal distance of 21.84 m, and launch angle of 45°. The correct time of flight was determined to be approximately 1.75 seconds for the vertical motion, leading to a horizontal velocity of 6.241 m/s. The initial vertical velocity equals the horizontal velocity, and the total initial velocity can be calculated using the Pythagorean theorem, resulting in a total initial velocity of approximately 8.81 m/s.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of vector components in physics
  • Basic algebra for solving equations
NEXT STEPS
  • Study the kinematic equation for vertical motion: d = 0.5gt²
  • Learn how to decompose vectors into horizontal and vertical components
  • Explore the Pythagorean theorem in the context of projectile motion
  • Investigate the effects of different launch angles on projectile trajectories
USEFUL FOR

Students and educators in physics, engineers working with projectile dynamics, and anyone looking to deepen their understanding of motion under gravity.

Otakuchica
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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
 
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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.
 
Otakuchica said:
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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