- #1
abdo99
- 2
- 0
Attempt solution
Y=(vsinø)t-0.5gt^2
4=(vsinø)t-4,9t^2
But there is 3 variables!
Another eq is
X=(vcosø)t
6=(vcosø)t
But i did not found any away to solution
Question in dynamics projectile motion
- Thread starter abdo99
- Start date
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The discussion centers on solving a projectile motion problem involving vertical and horizontal components of motion. The equations presented are Y = (v sin θ)t - 0.5gt² and X = (v cos θ)t, with specific values given for height and distance. Participants note the challenge of having three variables without sufficient information, particularly the maximum height. A hint suggests that knowing the maximum height can help calculate the initial velocity components. The rule for maximum height is provided as H = (u² sin² θ) / 2g, indicating a potential method for solving the problem.
Attempt solution
Y=(vsinø)t-0.5gt^2
4=(vsinø)t-4,9t^2
But there is 3 variables!
Another eq is
X=(vcosø)t
6=(vcosø)t
But i did not found any away to solution
- #2
dauto
- 1,948
- 201
You posted a sideways picture
- #3
rtsswmdktbmhw
- 38
- 2
Hint: 1. The ball travels on a parabolic path.
2. If you know the max height of the ball, you can calculate the vertical and horizontal components of initial velocity.
Or, at least, that's how I managed to do it but there may be a simpler method/formula..
2. If you know the max height of the ball, you can calculate the vertical and horizontal components of initial velocity.
Or, at least, that's how I managed to do it but there may be a simpler method/formula..
Last edited:
- #4
abdo99
- 2
- 0
I know that ,but i did not konw the maximum H
The rule is H=(u^2sin^2)/2g
The rule is H=(u^2sin^2)/2g
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