Man Throws Ball 138m Horizontally, How Far Vertically?

[a2k22]

A man can throw a ball a maximum horizontal
distance of 138 m.
The acceleration of gravity is 9.8 m/s2 .
How far can he throw the same ball vertically upward with the same initial speed?
Answer in units of m.

I figured out the initial speed to be about 280 m/s. Am I right? How can I finish the problem?

 

[Stovebolt]

A man can throw a ball a maximum horizontal
distance of 138 m.
The acceleration of gravity is 9.8 m/s2 .
How far can he throw the same ball vertically upward with the same initial speed?
Answer in units of m.

I figured out the initial speed to be about 280 m/s. Am I right? How can I finish the problem?

What steps did you take to determine the initial speed? I ran through the problem and got a much different answer.

 

[a2k22]

I used the equation range equals (initial velocity squared * sin of 2 theta) divided by gravity.
range equals 180
theta equals 45 (this is the angle that will achieve the farthest range)
and gravity equals 9.8 (the negative sign is not needed)

And I did the problem again, and I got about 1352

 

[nasu]

A good idea would be to show your solution.

 

[nasu]

And I did the problem again, and I got about 1352

You forgot to take the square root this time.
Is the distance 138 or 180?

 

[a2k22]

range is equal to 180, and yes, thank you. I always forget to square root. So it's about 36.77 m/s. Now how should I proceed?

 

[Stovebolt]

range is equal to 180, and yes, thank you. I always forget to square root. So it's about 36.77 m/s. Now how should I proceed?

Check the math on that one. It looks like you used 138 m for the range.

 

[a2k22]

I got 42 m/s. What now?

 

[Stovebolt]

Okay, thanks for catching my mistake.

But to be honest, I'm very tired. I would really like to go to bed. I try and figure out the problem tomorrow, I'll ask a classmate. But can I just get an answer please? I know you are trying to help me, and I appreciate it. It's just been a very tiring day, and I have to get up early tomorrow again.

Thank you.

Given the initial speed and the final speed (at max height, v = 0), use the formula
V_f² = V_i²+2ad

Solving for d and substituting knowns:

d = V_f²/(2 g) = 42²/(2*9.8) = 90 m

 

[a2k22]

It says the answer is wrong! (PS it's one of those online hw)