a ball player hits home run, and clears a wall 21 m high located 130 m from home plate. Hit at an angle of 35 degrees and is hit at a height of 1 m from the ground.
a) What is the initial speed of the ball?
I tried using the formula sq((R*g)/sin2(theta)), but I come out with an answer of 36.8, while the book says that the correct answer is 41.7 m/s. Am I using the wrong formula, or is the concept wrong?
Thanks!
2006-10-08 04:03:56 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
First draw the diagram.
then just take the horizontal componenet of velocity. so you will get time as t=130m/ucos35. now for the vertical component. using s=ut+0.5at^2. or
21=usin35(130/ucos35)+0.5(-9.8)*(130/ucos35)^2. now just solve for u
2006-10-08 05:47:00 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The problem with the formula you used is that it assumes a trajectory starting and ending at the same elevation. The baseball had to clear a 21 m wall.
You need to break the initial velocity into vertical and horizontal components: Vh = Vcos35, Vv = Vsin35. Analyse the vertical and horizontal travel independently but recognize that the flight time, t, is the same for both.
Vertical: use y = Vv*t - (1/2)*g*t^2 where y is the difference between the starting and ending elevation.
Horizontal: use x = Vh*t where x is 130 m. Solve this for t and substitute the result into the 1st equation.
Evaluate for your answer.
2006-10-08 07:32:32 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋