Exercise 3.17 A major leaguer hits a baseball so that it leaves the bat at a spe

Question

Exercise 3.17

A major leaguer hits a baseball so that it leaves the bat at a speed of 28.6 m/s and at an angle of 38.4 above the horizontal. You can ignore air resistance.

Part A

At what two times is the baseball at a height of 11.6 m above the point at which it left the bat?

Give your answers in ascending order separated with comma.

2.76,0.85

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Part B

Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part (a).

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Correct

Significant Figures Feedback: Your answer 22.41 m/s was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your answer.

Part C

Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

11.98

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Part D

Calculate the horizontal component of the baseball's velocity at a later time calculated in part (a).

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Part E

Calculate the vertical component of the baseball's velocity at a later time calculated in part (a).

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Part F

What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?

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Part G

What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

please i need the answeres with details.

Exercise 3.17

A major leaguer hits a baseball so that it leaves the bat at a speed of 28.6 m/s and at an angle of 38.4 above the horizontal. You can ignore air resistance.

Part A

At what two times is the baseball at a height of 11.6 m above the point at which it left the bat?

Give your answers in ascending order separated with comma.

t1,t2 =

2.76,0.85

  s  

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Incorrect; Try Again; 4 attempts remaining; no points deducted

Part B

Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part (a).

vx = 22.4   m/s  

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Correct

Significant Figures Feedback: Your answer 22.41 m/s was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your answer.

Part C

Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

vy =

11.98

  m/s  

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Incorrect; Try Again; 5 attempts remaining

Part D

Calculate the horizontal component of the baseball's velocity at a later time calculated in part (a).

vx =   m/s  

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Part E

Calculate the vertical component of the baseball's velocity at a later time calculated in part (a).

vy =   m/s  

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Part F

What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?

v =   m/s  

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Part G

What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

=    below the horizontal

please i need the answeres with details.

Explanation / Answer

The base ball leaves the bat at velocity 'V' =28.6m/s at an angle 38.4 degree 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 38.4 = 28.6*0.62=17.7 m/s
2.) Horizontal Component=V2= V cos38.4= 28.6*0.78=22.3 m/s

Note: the Horizontal component:V2 will be always Constant =22.3 m/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=17.7 m/s
h=11.6m, a= -g= -9.8m/s^2

h=vt+1/2 at^2
11.6=17.7t-4.9t^2
4.9t^2+11.6-17.7t=0
t=0.860 sec or 2.75 sec.

the ball reaches at height 11.6m in 0.860s (lowest) during accending in to air, so we can not take the accending time 2.75sec. This will occur during descending.

lets find the descending time at which the ball reaches the hieght 11.6m 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= 17.7*17.7/19.6=15.98 m
time taken to reach the Pick= V1/g=1.81 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=1.81-0.860=0.95


At the point of pick the ball is at constant Horizontal speed=22.3m/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-11.6=15.98-11.6= 4.38 from the pick.

so time(T) taken to fall 4.38m from pick.

4.38 =V3t+1/2 gT^2 =>4.38=0+4.9T^2 => T=0.95s

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 11.6m+ time taken to reach pick fro 11.6m height+ time taken to descend to height 10m during the projectile motion.
=0.860+0.95+0.95= 2.76 m
here you can calculate that " time taken to reach pick from 11.6m height = time taken to descend to height 11.6" during the projectile motion.

So answers are as under

a) t= 0.860 and 2.76 s at height 11.6 above the level at which it was hit by the bat.

Horizontal components of the velocity=22.3 m/s both the cases.
During ascending the vertical component: Vx = V1-gt=17.7-9.8*(0.860)=9.3 m/s
During descending: Vertical component= Vx=0.95×9.8 =9.3 m/s
So vertical componets will be 9.3m/s and Horizontal =22.3m/s at both the pints of ascending and descending.

The Magnitude of the velocity at the level it was hit by the ball is same as it was leaving the ball= 28.6m/s
and direction/angle at which it will reach there is 38.4 degree clock wise.

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