Baseball: Rising and Falling speed

[Trizz]

Homework Statement

A baseball is seen to pass upward by a window 28 m above the street with a vertical speed of 8 m/s. The ball was thrown from the street.

(a) What was its initial speed?
(b) What altitude does it reach?
(c) How long after it was thrown did it pass the window?
(d) After how many more seconds does it reach the street again?

Homework Equations

Vf = Vi + at
Vf^2 = Vi^2 + 2ad
d = Vi(t) + (.5)(a)(t)^2
t = Vf/a or Vf-Vi/a

The Attempt at a Solution

So far I've only attempted a and c because I don't know how to go aout solving b or d.

For a, I used the Vf^2 formula and changed it to equal Vi^2. So I had:

2(a)(d) - Vf^2 = Vi^2 and with that I got 2(9.8)(28) - (8)^2 = Vi^2 and got 22

I believe the problem may be with acceleration. Is it 9.8 when thrown uo, or only during free fall?


For c, I took the answer i got from a, Vi = 22, and plugged it into the time equation. I did 8 - 22 / 9.8 to get 1.4 seconds.

 

[Nate810]

I'm not sure if my answers for a and c are correct. If someone could help me with b and d, I'd really appreciate it.

 

[tomcue]

I would like to clarify that the acceleration due to gravity, represented by 'a' in the equations, remains constant at 9.8 m/s^2 throughout the motion of the baseball. This is because the force of gravity acts on the ball continuously, causing it to accelerate downward at a constant rate. Therefore, in the equation Vf^2 = Vi^2 + 2ad, 'a' should always be 9.8 m/s^2.

Using this information, I would like to provide a response to the homework questions:

(a) To find the initial speed of the baseball, we can use the equation Vf = Vi + at. We know that the final speed (Vf) is 8 m/s, and the time (t) is 1.4 seconds (as calculated in part c). Therefore, we can rearrange the equation to solve for Vi, which gives us an initial speed of 19.6 m/s.

(b) To calculate the altitude that the ball reaches, we can use the equation d = Vi(t) + (0.5)(a)(t)^2. We know that the initial speed (Vi) is 19.6 m/s and the time (t) is 1.4 seconds. Plugging these values into the equation, we get a maximum altitude of 19.6 meters.

(c) We can use the same equation as in part b to calculate the time it takes for the baseball to pass the window. We already know the initial speed (Vi) and the acceleration (a), so we can rearrange the equation to solve for time (t). This gives us a time of 1.4 seconds, which is the same as the time calculated in part c.

(d) To determine the time it takes for the ball to reach the street again, we can use the equation t = Vf/a. We know that the final speed (Vf) is 8 m/s and the acceleration (a) is 9.8 m/s^2. Plugging these values into the equation, we get a time of 0.82 seconds. Therefore, it takes a total of 1.4 + 0.82 = 2.22 seconds for the ball to reach the street again.

I hope this helps to clarify your understanding and approach to solving the problem. Remember to always consider the constant acceleration due to gravity when