View Full Version : More physics.... Help Please!!
Jameson
Nov28-04, 08:32 PM
A ball player hits a home run, and the baseball just clears a wall 18.5 m high located 140.0 m from home plate. The ball is hit at an angle of 32° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.0 m above the ground.
a. What is the initial speed of the ball? Solve for time in the x-direction and substitute it in the formula for the vertical position.
Incorrect. Tries 1/15
b. How much time does it take for the ball to reach the wall? Tries 0/15
c. Find the velocity components and the speed of the ball when it reaches the wall.
Vy,f
Vx,f
Vf
-----------------
I have no idea how to start :yuck:
donjennix
Nov28-04, 08:46 PM
Well let's start with what we know --
We seem to be told to assume that the 18.5 m wall is the apex of the ball's height and the ball started at 1 m so it travelled 17.5 m up before its vertical vel went to zero (vertical velocity is zero when the ball stops rising and starts falling). Velocity after time t is initial vertical velocity (Vy here I think) - gt -- that is:
V(17.5 m) = Vy - gt = Vy - 9.8t
If we set that to 0 we get that
Vy = 9.8t -------- first step
We also know that the distance up (17.5 m) is as follows
17.5 = Vyt - gt^2/2 --- But note that we have an expression for Vy just above so
17.5 = 9.8t^2 - 9.8t^2/2 = 4.9t^2
t = SQRT(17.5/4.9) = 1.89 sec
So Vy when the ball left the bat was:
9.8 * 1.89 = 18.522 m/sec
I did this straight on the computer and I have not checked whether I made any errors -- so check it -- the logic should be right.
Why don't I let you work the rest.
Jameson
Nov28-04, 08:55 PM
Thank you for your help, but my online homework says that is the wrong answer. So if someone could still help me, I'd appreciate it.
donjennix
Nov29-04, 10:25 AM
OK - one more try; I appear to have been wrong in assuming that 18.5 m was the apex of the ball's flight. Using the hint from part a.,
where t is the time at which the ball reached the fence
travel in x direction:
140 m = Vx*t so
t=140/Vx
travel in y direction:
17.5 = Vy*t - 9.8*t^2/2
17.5 = (Vy/Vx)*140 - 0.5*9.8*19600/Vx^2 (equation 1)
I think we assume tan(32)= Vy/Vx
i.e., 0.6249 = Vy/Vx or Vy = 0.62489*Vx
Using this to solve equation 1 above yields Vx= 37.045 m/sec so
Vy = .6249*37.045 = 23.14 m/sec.
For "initial speed" they probably want the velocity at 32 deg from the horiz so that would be Vy/cos(32) or Vx/sin(32)
Well, hope this one was right
I'll leave b) for you
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