A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 34.9 degrees above the horizontal. You can ignore air resistance. A) At what two times is the baseball at a height of 10.0 above the point at which it left the bat? B) Calculate the horizontal component of the baseball's velocity at each of the two times you found in part A. C) Calculate the vertical component of the baseball's velocity at each of the two times you found in part A D) What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat? E) What is the direction (in degrees below the horizontal) of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

there will be two components of 'V' Vertical component =V1 = V sin 36.9= 30 * 0.6=18 m/s Horizontal Component=V2= V cos36.9=30 *0.8=24 m/s Initial velocity of Ball vertically=18m/s h=vt+1/2 at^2 9=18t-4.9t^2 4.9t^2+9-18t=0 t=0.597 or 3.08 sec. the ball reaches at height 9m in 0.682s (lowest) during accending in to air, so we can not take the accending time 2.99sec. This will occur during descending. The ball will reach a point where the vertical component will be =0 ie V3 at the top of the projectile=0 so total height covered by ball. H=(V3^2-V1^2)/2a = -V1^2/ -2g= 18*18/19.6=16.53 m time taken to reach the Pick= V1/g=1.84 s from the level at which the bat hit the ball. So the time taken to reach the pick from 9m height from the level at which the bat hit the ball=1.84-0.597= 1243 At the point of pick the ball is at constant Horizontal speed=24m/s But the gravity will pull it down, that is why the ball will descend with a curve same as it was ascending in to the air. At pick the ball's initial downward speed U=V3=0. To reach at height 10m above the level at which it was hit by the bat, the ball will fall=H-10=16.53-9= 7.53from the pick. so time(T) taken to fall 7.53m from pick. 7.53=V3t+1/2 gT^2 =>7.53=0+4.9T^2 => T=1.24s So total time taken to reach at height 9m above the level at which it was hit by the bat during descending =Time taken to ascend upto 9m+ time taken to reach pick fro 9m height+ time taken to descend to height 9m during the projectile motion. =0.597+1.24+1.24 = 3.077 here you can calculate that " time taken to reach pick from 9m height = time taken to descend to height 9m during the projectile motion. So answers are as under a) t= 0.597s and 3.08s at height 9m above the level at which it was hit by the bat. b)Horizontal components of the velocity=24m/s both the cases. During ascending the vertical component: Vx = V1-gt=18-9.8*(0.597)=12.15 m/s During descending: Vertical component= Vx=1.24*9.8 =12.15m/s So vertical componets will be 12.15m/s and Horizontal =24m/s at both the pints of ascending and descending. c) the Magnitude of the velocity at the level it was hit by the ball is same as it was leaving the ball= 30m/s and direction/angle at which it will reach there is 36.9 degree clock wise.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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