Calculating Initial Velocity of Projectile Motion Without Time Information

[rl.bhat]

Your formula for y is wrong. It should be
yf = yi + vt - 1/2*g*t^2.

 

[physicskid69]

Your formula for y is wrong. It should be
yf = yi + vt - 1/2*g*t^2.

I let g=-9.8m/s^2 in my calculations.

 

[rl.bhat]

In the equation what is the value of yi?
After the bounce it should be zero. Take it zero and calculated the time.

 

[physicskid69]

I now use that velocity with yf=yi+vt+1/2gt^2 to solve for t where yf=0.1m and yi=0m. This gives me t=0.881s.

Isn't that what I already did?

 

[rl.bhat]

Yes. I didn't notice.
0.1 = 4.43*t - 4.9*t^2
Or 4.9*t^2 - 4.43*t + 0.1 = 0
When I solved it I got different answer. Check it.

 

[physicskid69]

Yes. I didn't notice.
0.1 = 4.43*t - 4.9*t^2
Or 4.9*t^2 - 4.43*t + 0.1 = 0
When I solved it I got different answer. Check it.

I get t=0.023s or t=0.881s and since I want the time for when the ball is coming back down I use t=0.881s I believe.

 

[rl.bhat]

OK.Then your answer is correct.

 

[physicskid69]

Okay, thank you to everyone that helped out with this.

 

[rl.bhat]

Wait a minute. There is one flaw in your solution.
According to you the total time of flight is 1.333 s.
During that time the vertical displacement is 8.7 m!
How is that? .

 

[physicskid69]

That's assuming the ball is falling the entire time and not considering the bounce.

 

[rl.bhat]

You have taken two time intervals to calculate the time of flight, taking into account the bounce.
But the vertical space in only 1m!

 

[physicskid69]

You have taken two time intervals to calculate the time of flight, taking into account the bounce.
But the vertical space in only 1m!

Yes I know, the ball falls 1m then bounces back up 1m and finally lands after traveling another 0.9m. This takes a total time of 1.333s.

 

[rl.bhat]

OK. Then it is alright.