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

[UrbanXrisis]

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

 

[vsage]

d=.5at^2
40m=.5(9.8)(t^2)
t=2.857142857 s

Since the ball was kicked at an initial velocity v_1, d = v_1t + \frac {1}{2}at^2
You're solving for v_1 eventually. The calculation you made for v after that was the average velocity which isn't entirely right.

 

[UrbanXrisis]

shoot, you're right

how should I go about solving this?

 

[UrbanXrisis]

Since the ball was kicked at an initial velocity v_1, d = v_1t + \frac {1}{2}at^2
You're solving for v_1 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.

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.

 

[Doc Al]

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.

 

[Skomatth]

I got 9.9 also.

 

[Galileo]

Since the ball was kicked at an initial velocity v_1, d = v_1t + \frac {1}{2}at^2

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.

 

[vsage]

Sorry I totally didn't see that it was horizontally kicked just that it was kicked.