How Is the Initial Velocity of a Horizontally Kicked Ball Calculated?

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Homework Help Overview

The problem involves a ball kicked horizontally off a 40m high cliff, with the goal of calculating its initial velocity based on the time it takes for the ball to hit the ground and the sound of the impact to travel back to the observer at the top of the cliff.

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  • Mixed

Approaches and Questions Raised

  • Participants discuss the time it takes for the ball to fall and the sound to travel back, exploring kinematic equations and the relationship between horizontal and vertical motion.

Discussion Status

Some participants have provided feedback on calculations, with a few expressing agreement on the initial velocity found. There is an ongoing exploration of the correct application of kinematic equations and the distinction between average velocity and initial velocity.

Contextual Notes

Participants are navigating the complexities of horizontal versus vertical motion, with some confusion about the initial conditions and the implications of the problem setup.

UrbanXrisis
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A ball is kicked horizontally off a cliff 40m high. After 3seconds, the person that is on top of the cliff hears the thump of the ball hitting the floor. Assuming speed of sound travles 343m/s in air, what is the initial velocity that the ball was kicked at?

Please check if I did it correct...

Find time it takes for the ball to hit the floor:
d=.5at^2
40m=.5(9.8)(t^2)
t=2.857142857 s

Find time it takes for the sound to travel back to person on cliff:
3-2.857142857=.1428571429

Find the distance away from cliff:
d=vt
d=343*.1428571429
d=49

Find the distance travled in the x-direction:
49^2=40^2+x^2
x=28.3019434m

Find the initial speed:
v=d/t
v=28.3019434m/2.857142857s
v=9.9m/s
 
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d=.5at^2
40m=.5(9.8)(t^2)
t=2.857142857 s
Since the ball was kicked at an initial velocity [tex]v_1[/tex], [tex]d = v_1t + \frac {1}{2}at^2[/tex]
You're solving for [tex]v_1[/tex] eventually. The calculation you made for v after that was the average velocity which isn't entirely right.
 
shoot, you're right

how should I go about solving this?
 
vsage said:
Since the ball was kicked at an initial velocity [tex]v_1[/tex], [tex]d = v_1t + \frac {1}{2}at^2[/tex]
You're solving for [tex]v_1[/tex] eventually. The calculation you made for v after that was the average velocity which isn't entirely right.

I was solving for the time it took the ball to drop. It has no initial velocity.

vsage said:
The easiest thing to do in the question is solve for how much time it takes for the sound to travel back up. since t = d/v you know how much time it took for the sound to move 40m. The total time for the events in the question to occur was 3 seconds. How much time did the ball spend falling then? You now have only one unknown in the kinematics equation I gave above. I hope this helps.

That's exactly what I did.
 
UrbanXrisis said:
A ball is kicked horizontally off a cliff 40m high. After 3seconds, the person that is on top of the cliff hears the thump of the ball hitting the floor. Assuming speed of sound travles 343m/s in air, what is the initial velocity that the ball was kicked at?

Please check if I did it correct...
Looks good to me.
 
I got 9.9 also.
 
vsage said:
Since the ball was kicked at an initial velocity [tex]v_1[/tex], [tex]d = v_1t + \frac {1}{2}at^2[/tex]
The initial velocity was in the x-direction, while the acceleration is in the y-direction. So this isn't the way to get the distance d.

UrbanXrisis' solution looks good to me.
 
Sorry I totally didn't see that it was horizontally kicked just that it was kicked.
 

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