Velocity Calculation of Kicked Rugby Ball

AI Thread Summary
The discussion revolves around calculating the velocity of a rugby ball kicked vertically upwards at an initial speed of 16 m/s after 3 seconds. Participants explore various equations and concepts, including the effects of gravitational acceleration acting against the ball's upward motion. Initial attempts yield incorrect results due to misunderstanding the direction of acceleration and the need to account for the ball's peak height. Ultimately, the correct approach involves recognizing that the ball will reach a maximum height before descending, leading to a final velocity calculation. The final accepted answer reflects the complexities of the physics involved in projectile motion.
Rumplestiltskin
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Homework Statement


What is the velocity of a rugby ball 3.0s after it is kicked vertically upwards with a speed of 16m/s? Give your answer to 2 s.f.

Homework Equations


SUVAT

The Attempt at a Solution


s = ?
u = 16
v = ?
a = g = 9.8
t = 3

Erm... v = u + at?
16 + 9.8 x 3 = 45m/s. Incorrect, try again.
...aha! Ball was kicked vertically up. Acceleration is going to be acting against. Not today, physics!
16 - 9.8 x 3 = -13m/s. Incorrect, try again.
Huh. Well, it might be a case of direction. Let's try plain 13m/s. Incorrect, try again.
Okay, let's work out the displacement so we can try v = √2gs. s = 16 x 3 - 0.5 x 9.8 x 3^2 = 3.9m. Nice round number! Must be on to something.
v = √2 x 9.8 x 3.9 = 8.7m/s. Incorrect, try again.
Wait, I'm a doughnut. That's just for free fall. I have a u value. So it's going to be v = √u^2 + 2as.
√16^2 + 2 x 9.8 x 3.9 = 18m/s. Incorrect, try again.
Uh. Oh, g would still be negative. This has got to do it.
√16^2 - 2 x 9.8 x 3.9 = 13m/s. I'm done. SOS.
 
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Rumplestiltskin said:
...aha! Ball was kicked vertically up. Acceleration is going to be acting against. Not today, physics!
16 - 9.8 x 3 = -13m/s. Incorrect, try again.
Looks right to me.
 
Rumplestiltskin said:

Homework Statement


What is the velocity of a rugby ball 3.0s after it is kicked vertically upwards with a speed of 16m/s? Give your answer to 2 s.f.

Homework Equations


SUVAT

The Attempt at a Solution


s = ?
u = 16
v = ?
a = g = 9.8
t = 3

Erm... v = u + at?
16 + 9.8 x 3 = 45m/s. Incorrect, try again.
...aha! Ball was kicked vertically up. Acceleration is going to be acting against. Not today, physics!
16 - 9.8 x 3 = -13m/s. Incorrect, try again.
Huh. Well, it might be a case of direction. Let's try plain 13m/s. Incorrect, try again.
Okay, let's work out the displacement so we can try v = √2gs. s = 16 x 3 - 0.5 x 9.8 x 3^2 = 3.9m. Nice round number! Must be on to something.
v = √2 x 9.8 x 3.9 = 8.7m/s. Incorrect, try again.
Wait, I'm a doughnut. That's just for free fall. I have a u value. So it's going to be v = √u^2 + 2as.
√16^2 + 2 x 9.8 x 3.9 = 18m/s. Incorrect, try again.
Uh. Oh, g would still be negative. This has got to do it.
√16^2 - 2 x 9.8 x 3.9 = 13m/s. I'm done. SOS.
This is what happens when you throw a bunch of formulas at a problem before thinking it through.

Even rugby balls kicked vertically upward can only rise so high.

What happens when the kicked ball goes up and then starts to come back down within this 3 second interval?
 
SteamKing said:
This is what happens when you throw a bunch of formulas at a problem before thinking it through.

Even rugby balls kicked vertically upward can only rise so high.

What happens when the kicked ball goes up and then starts to come back down within this 3 second interval?

Noted.

The velocity reaches 0, and then rises again. How do I tell what point it's at? Could still be rising.
 
Rumplestiltskin said:
The velocity reaches 0, and then rises again. How do I tell what point it's at?
You don't need to where this occurs, just when it occurs.
 
haruspex said:
Looks right to me.

Huh... just retried and it was accepted this time. That was a headache. Thanks!
 
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