A major leaguer hits a baseball so that it leaves the bat at a speed of 30.5m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.5m/s and at an angle of 34.9 above the horizontal. You can ignore air resistance.

a) At what two times is the baseball at a height of 8.00m 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 of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

The base ball leaves the bat at velocity 'V' =30.5 m/s at an angle 36.9degree with horizontal at the level at which it left the bat.

So there will be two components of 'V'
1.) Vertical component =V1 = V sin 34.9= 30.5 * 0.6=17.45 m/s
2.) Horizontal Component=V2= V cos34.9=30.5 *0.8=25.01 m/s

Note: the Horizontal component:V2 will be always Constant =25m/s as there is no acceleration/deceleration towards horizontal movement.


The vertical Component will take the ball to a height vertically.

Initial velocity of Ball vertically=17.45m/s
h=8m, a= -g= -9.8m/s^2

h=vt+1/2 at^2
8=17.45t-4.9t^2

t=0.54 or 3.02 sec.

the ball reaches at height 8m in 0.54s (lowest) during accending in to air, so we can not take the accending time 3.02 sec. This will occur during descending. Let's prove it.

lets find the descending time at which the ball reaches the hieght 10m above the level at which it was hit by the bat.

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 10m height from the level at which the bat hit the ball=1.84-0.682= 1.15


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=16.53-10= 6.53from the pick.

so time(T) taken to fall 6.53m from pick.

6.53=V3t+1/2 gT^2 =>6.53=0+4.9T^2 => T=1.15s

So total time taken to reach at height 10m above the level at which it was hit by the bat during descending
=Time taken to ascend upto 10m+ time taken to reach pick fro 10m height+ time taken to descend to height 10m during the projectile motion.
=0.682+1.15+1.15 = 2.99
here you can calculate that " time taken to reach pick from 10m height = time taken to descend to height 10m" during the projectile motion.

So answers are as under

a) t= 0.682s and 2.99s at height 10m 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*(0.682)=11.3 m/s
During descending: Vertical component= Vx=1.15*9.8 =11.3m/s
So vertical componets will be 11.3m/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.

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