- #1
mooha
- 5
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Not a literal twist! It's just a bit more complex.
A ball is to be shot from level ground toward a wall at a distance "x". I am given the graph of the y component of velocity just as it would reach the wall, as a function of that distance x(m). Here's the equation of the line:y=-.5x+5, [0, 20]
What is the launch angle?
v = vo +at
x = xo + Vot + .5at2
v2 = (vo)2 + 2a(x - xo)
So I wrote out all I knew about the x and y components
y: vo = 5, a = -9.8, xo = 0
x: a=0
I attempted to created a triangle of the horizantal and vertical components of velocity and the initial velocity as the hypotenuse solve with trig. I got
sin(θ)= (5)/(V)
tan(θ)= (5)/(vox)
No help.
I really am stuck! Using kinematics, you can't find the initial velocity because you don't know the time. You can't find the time, because you don't know where in the 20 m of the trajectory the ball hit the wall (according to my interpretation of the problem) Ah! If you could help I would be so grateful!
Thank you so much!
Homework Statement
A ball is to be shot from level ground toward a wall at a distance "x". I am given the graph of the y component of velocity just as it would reach the wall, as a function of that distance x(m). Here's the equation of the line:y=-.5x+5, [0, 20]
What is the launch angle?
Homework Equations
v = vo +at
x = xo + Vot + .5at2
v2 = (vo)2 + 2a(x - xo)
The Attempt at a Solution
So I wrote out all I knew about the x and y components
y: vo = 5, a = -9.8, xo = 0
x: a=0
I attempted to created a triangle of the horizantal and vertical components of velocity and the initial velocity as the hypotenuse solve with trig. I got
sin(θ)= (5)/(V)
tan(θ)= (5)/(vox)
No help.
I really am stuck! Using kinematics, you can't find the initial velocity because you don't know the time. You can't find the time, because you don't know where in the 20 m of the trajectory the ball hit the wall (according to my interpretation of the problem) Ah! If you could help I would be so grateful!
Thank you so much!
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