In the figure, a baseball is hit at a height h = 1.20 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.2 s after it is hit and then down past the top of the wall 4.1 s later, at distance D = 55 m farther along the wall. (a) What horizontal distance is traveled by the ball from hit to catch? What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is the wall?
For part A I'm using the following equations
The Attempt at a Solution
I have been working this problem for over an two hours now and even though it seems simple i keep coming up with the wrong answers.
So, first i know that i have to split up the problem into two parts:
there should be no acceleration in the x direction, so at constant velocity
the vx component should be simply distance traveled/ time.
I use the section that states the distance between 1.2 and 4.1 seconds as 55m so t=4.1-1.2=2.9s → 55m/2.9s=18.97m/s. This is the vx component of the initial velocity vector.
I use this equation: rf=r0+v0t+1/2at2
rearranging for vnot=-1/2at2/t.
I know a=-9.8m/s, t will be the sum of the sections of time so 1.2s + 2.9s + 1.2s = 5.3s
the Δr should be zero, since it will end up falling to the same height that it left from.
Plugging in i got 25.97m/s.
(B) The initial velocity vector should simply be the sqr(v0x2+v0y2) which gives me 32.1606 m/s
(C)To find the angle of the initial velocity vector i take the inverse tangent of the component vector to obtain, 53.85 °
(D) To find the height of the wall i simply add 1.2m to the result from plugging in 1.2 seconds (the point at which it clears the wall) into the y component equation and using the ynot comp vector and solving for Δr which gives me (1.2m+31.164m-7.056m)=25.308m
What am i doing wrong? it seems to me I'm doing every single step right but then again I'm suffering from insomnia so maybe I'm just missing one crucial step and i don't realize it?