I launch a ball at a height of .24 meters with a velocity of 7.5m/s. The launch
ID: 2184691 • Letter: I
Question
I launch a ball at a height of .24 meters with a velocity of 7.5m/s. The launch angle is 57 degrees and the ball falls to the ground.What would be the second angle the projectile can be launched and reach the same range?
I'm not sure how to solve this problem. R=u^2sin(2theta)/g
However, this equation only applies in situations where the projectile lands at the same elevation from which it was fired.
So how would i solve a problem where the projectile lands at a DIFFERENT elevation from which it was fired?
Explanation / Answer
I like to just break it up into motion in the y and x directions.
For the first ball,
Initial Position:
Yi = .24 m
Xi = 0 m
Initial Velocity:
Vyi = 7.5sin(57)
Vxi = 7.5cos(57)
Acceleration:
Ay = -9.8 m/s^2
Ax = 0
Now use:
Yf = Yi + Vyi*t + (1/2)(-9.8)t^2
to solve for t
0 = .24 + 7.5sin(57)*t + (1/2)(-9.8)t^2
t=1.32076 s
Now use this t to solve for the range (displacement in the x direction)
Xf = Xi + Vxi*t
Xf = 0 + 7.5cos(57)(1.32076)
Xf = 5.395 m
Now for the second angle:
We know that:
5.395 = Xf = 7.5cos()t
We also know that t is a solution of the equation:
0 = .24 + 7.5sin()*t + (1/2)(-9.8)t^2
So let's solve the first equation for t:
t = 5.395/(7.5cos)
And plug this value into the second equation:
0 = .24 + 7.5sin()*(5.395/(7.5cos)) + (1/2)(-9.8)( 5.395/(7.5cos))^2
Now we can solve this equation for the angle (which we know is between 0 and 90:
(I had a numerical solver (wolframalpha) do it because it looked ugly)
This gives 2 angle between 0 and 90
57 (as we would expect)
and
30.452
You can go thorugh and check the range with that angle, but I am sure that we did it correctly.