Physics kinematics problem (not homework, but my own)

[grooveactiva]

Homework Statement

This is a physics problem that I made up in order to understand one aspect of kinematics in 1-dimension: A baseball is thrown straight up, leaving the hand which is 1.1 meter above the ground. It reaches a height of 12 meters. (a) What must be the speed of the baseball when it leaves the hand to reach this height? (b) How long does it take for the baseball to fall to the ground? (c) At what speed does it have just before it strikes the ground?

Homework Equations

y = y_i + v_i(Δt) + ½a(Δt)²
v_f - v_i = a(Δt)
v_f² - v_i² = 2a(Δs)
a =g = 9.8 m/s²

The Attempt at a Solution

(a) How fast baseball leaves hand:
v_f² - v_i² = 2(Δt)
v_i = 0 m/s
v_f² + 0 = 2(-9.8 m/s²)(12-1.1 m)
v_f = √(2*-9.8*10.9) = 14.6 m/s

(b) How long (how much time) does it take for baseball to fall to the ground:
v_f - v_i = a(Δt)
v_f = 0 m/s because ball hits ground
(Δt) = (0 - 14.6 m/s)/-9.8 m/s² = 1.5 seconds
Or, should I have used y = y_i + v_i(Δt) + ½a(Δt)², with v_f = 14.6 m/s
But then I get via the quadratic formula: time = -0.0074 s (discard because negative time) or 3.05 seconds
Why doesn't these two equations match in their anwers for time: 1.5 seconds vs. 3.05 seconds?

(c) At what speed does the baseball have just before it strikes the ground: Not sure how to proceed until I can figure out the answer to part (b) above.

 

[Orodruin]

vf = 0 m/s because ball hits ground

This is wrong. It violates your basic assumption that the acceleration of the ball is just due to gravity and the resulting parabolic motion. In order to stop the ball, you need a force from the ground. Instead, you need to look for when the ball has the appropriate displacement.

 

[grooveactiva]

Okay. So, the velocity of the baseball when it hits the ground is not zero.
This makes sense because the ball bounces, so there is a force opposing the fall of the baseball. So it must be 3.05 seconds.
But now I am confused: Is this 3.05 seconds the amount of time the ball is in the air from the time it was thrown up (from height of 1.1 m), then it reaches its peak height, and then falls to the ground? Or is it just the time it takes for it to fall from peak height to the ground?
I am assuming the former because the equation of displacement has a parabolic path to it (Δt)².

 

[PeroK]

Okay. So, the velocity of the baseball when it hits the ground is not zero.
This makes sense because the ball bounces, so there is a force opposing the fall of the baseball. So it must be 3.05 seconds.
But now I am confused: Is this 3.05 seconds the amount of time the ball is in the air from the time it was thrown up (from height of 1.1 m), then it reaches its peak height, and then falls to the ground? Or is it just the time it takes for it to fall from peak height to the ground?
I am assuming the former because the equation of displacement has a parabolic path to it (Δt)².

There's a quick way to estimate how far a ball will fall in 3s. The final speed is about 30m/s, hence the average speed is 15m/s. Times 3s is 45m.

Therefore, as you suspected, 3s is enough time for the ball to go up and down about 12m.

 

[grooveactiva]

So, now I am confused again. If I wanted to find out how fast the baseball is going just _before_ it hits the ground, it can be calculated by now using the v_f = v_i + aΔt, but the t is for the entire flight (from 1.1 m up to 12 meters, then down to 0 meters ground), not just the time to fall from a height of 12 m.
So, I use v_f² - v_i² = 2a(Δs): v_f² = 0 m/s + 2(-9.8)(12m) = 15.336 or 15 m/s
But what if I wanted to find how long it took the ball to fall downward only from the height of 12 m (and not just the entire parabolic flight path up and down):
Can I then use v_f = v_i + aΔt, with v_f = 15.336 m/s and v_i = 0 m/s and then solve for Δt? To get Δt = 1.56 s.

 

[PeroK]

So, now I am confused again. If I wanted to find out how fast the baseball is going just _before_ it hits the ground, it can be calculated by now using the v_f = v_i + aΔt, but the t is for the entire flight (from 1.1 m up to 12 meters, then down to 0 meters ground), not just the time to fall from a height of 12 m.
So, I use v_f² - v_i² = 2a(Δs): v_f² = 0 m/s + 2(-9.8)(12m) = 15.336 or 15 m/s
But what if I wanted to find how long it took the ball to fall downward only from the height of 12 m (and not just the entire parabolic flight path up and down):
Can I then use v_f = v_i + aΔt, with v_f = 15.336 m/s and v_i = 0 m/s and then solve for Δt? To get Δt = 1.56 s.

That all looks about right.