In baseball, there are times when the batter makes contact with the ball only to

Question

In baseball, there are times when the batter makes contact with the ball only to see it pop straight up in the air above home plate. Suppose we have a situation where the batter makes contact with the ball at 1.50m above home plate, and the ball hits off the bat at 90.0 mph (miles per hour).

a. Suppose the catcher catches the ball at the same height the ball hit off the player's bat. How long does it take for the ball to reach the catcher's glove?

b. Suppose the catcher misses the ball. How long until the ball reaches the ground?

Explanation / Answer

As the ball pop straight up in the air above home plate, the case resembles the case of vertically upward projectile with initial velocity The velocity of the gall is u = 90 mph                                          = (90)(0.45 m/s)                                          = 40.5 m/s The height of the batter above the ground is h = 1.5 m a) As the catcher catches the ball at the same height as projected, the time spent by the ball in the air is called time of flight and is given by             T = 2u/g                = 2(40.5 m/s)/(9.8 m/s2)                = 8.26 s The ball spents a time of 8.26 s before the catcher caught the ball b) Supose the catcher misses the ball Then the ball will have same velocity at the point of projection i.e. 1.5 m from the home plate but in the opposite direction This is the case of an object falling from height 1.5 m with velocity 40.5 m/s The equation of motion is given by h = ut + 0.5gt2 1.5 m = (40.5 m/s)t + 0.5(9.8 m/s2)t2 4.9t2 + 40.5t - 1.5 = 0 t = 0.037 s The total time spent by the ball in the air is T = 8.26 s + 0.037 s                                                                    = 8.3 s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.