A Student Throws A Baseball Vertically Upward?
Contents
- Introduction
- The student throws the ball with some speed v0 upward.
- The ball reaches some height h after some time t.
- The ball’s speed when it reaches height h.
- The ball’s speed when it reaches the top of its trajectory.
- The ball’s speed when it reaches the ground again.
- The ball’s speed when it reaches the student’s hand again.
- Conclusion
A student throws a baseball vertically upward with an initial speed of 20.0 m/s. At what height will the baseball be caught by the catcher?
Introduction
In this experiment, you will study the motion of a baseball thrown upward. You will use the computer to help you analyze the data and graph the motion.
The student throws the ball with some speed v0 upward.
At the instant the student releases the ball, it is not moving horizontally, so its initial horizontal velocity is zero. For the ball to reach the height h of the bleachers, it must rise vertically a total distance of h + y0. Because the initial vertical velocity is v0 > 0, gravity will cause the ball to start falling downward as soon as it is released.
The ball reaches some height h after some time t.
A student throws a baseball vertically upward with an initial velocity of+10 ft/s. The ball reaches some height h after some time t.
-h=(10t-16t^2)ft
-At what height does the ball reach its maximum height?
-How long does it take the ball to reach its maximum height?
-What is the velocity of the ball at its maximum height?
-What is the velocity of the ball at t=4s ?
-How long does it take the ball to hit the ground again?
-With what velocity does it hit the ground?
The ball’s speed when it reaches height h.
At the height h, the ball’s speed is given by:
v(h) = v0 + gt – gh
where v0 is the speed with which the ball was thrown upwards, g is the acceleration due to gravity, and t is the time taken for the ball to reach height h. Solving for t in terms of v0 and h gives:
t = (v0 + sqrt(v0^2 + 2gh))/g
Substituting this value of t back into the original equation gives:
v(h) = v0 + gt – gh = v0 + g((v0 + sqrt(v0^2 + 2gh))/g) – gh = v0 + (v0 + sqrt(v0^2 + 2gh)) – gh = sqrt(v0^2 + 2gh) – gh
The ball’s speed when it reaches the top of its trajectory.
What is the ball’s speed when it reaches the top of its trajectory?
Assuming there is no air resistance, the ball’s speed when it reaches the top of its trajectory would be the same as its initial speed.
The ball’s speed when it reaches the ground again.
Assuming there is no air resistance, the ball’s speed when it reaches the ground again will be the same as its speed when it was thrown upward.
The ball’s speed when it reaches the student’s hand again.
The ball’s speed when it reaches the student’s hand again is zero.
Conclusion
After reading and doing the calculations, we have found that a student throws a baseball vertically upward, it will take approximately 3.5 seconds for the baseball to reach its peak. The student will then have to wait another 3.5 seconds for the baseball to come back down to them. If the student were to catch the baseball at the very peak of its arc, they would need excellent timing and coordination.
