What is the speed of the pitch?

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sjb-2812 said:
Are you sure R should be 25 here?

I'm never sure, 8^)

What do you think it should be?
 
Your "kinematic equations" refer to the vertical motion only. You say the ball is thrown horizontally so the initial vertical component is 0. The height of the ball, at time t, is [itex]y= (1/2)(9.81)t^2+ 4[/itex]. Set that equal to 0 (the ball hits the ground) and solve for t to find when the ball hits the ground.

Taking "v" as the horizontal component of velocity, the distance the ball traveled, in time t, is given by x= vt. Put x= 25, t equal to the time you solved for above, and solve for v.

That, of course, is ignoring air resistance- which is fairly large for a baseball.
 
Spinnor said:
I'm never sure, 8^)

What do you think it should be?

Maybe it's the way the question is worded, but the figure of 25 to me seems to be the hypotenuse, rather than the adjacent for the triangle?


HallsofIvy said:
Your "kinematic equations" refer to the vertical motion only. You say the ball is thrown horizontally so the initial vertical component is 0. The height of the ball, at time t, is [itex]y= (1/2)(9.81)t^2+ 4[/itex]. Set that equal to 0 (the ball hits the ground) and solve for t to find when the ball hits the ground.

Taking "v" as the horizontal component of velocity, the distance the ball traveled, in time t, is given by x= vt. Put x= 25, t equal to the time you solved for above, and solve for v.

That, of course, is ignoring air resistance- which is fairly large for a baseball.

Check your signs here :)
 
You wrote,

"Maybe it's the way the question is worded, but the figure of 25 to me seems to be the hypotenuse, rather than the adjacent for the triangle?"

On second reading I agree, thanks for the correction.