The discussion centers on calculating the pitching speed of a baseball thrown horizontally from a height of 4.0 meters, landing 25 meters away. Participants utilize kinematic equations to derive the time of flight based on vertical motion, specifically using the equation y = (1/2)(9.81)t² + 4. The horizontal distance is then expressed as x = vt, where v represents the horizontal velocity. The conversation highlights the importance of correctly interpreting the distance as the horizontal component and addresses the impact of air resistance on the calculation.
PREREQUISITESPhysics students, educators, and sports scientists interested in understanding the dynamics of projectile motion and its applications in sports like baseball.
Spinnor said:See towards the bottom of,
http://ngsir.netfirms.com/englishhtm/ThrowABall.htm
Set h = 4, R = 25 , and θ = 0
sjb-2812 said:Are you sure R should be 25 here?
Spinnor said:I'm never sure, 8^)
What do you think it should be?
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 y= (1/2)(9.81)t^2+ 4. 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.