Question

A baseball traveling upward passes a window 29.0 m above the street with a vertical speed of 14.0 m/s. If the ball was thrown from street level, what is the maximum height that the baseball reaches?
Im stumped, please tell me what equations to use as well as what each variable in the equation is calling for

Answer 1

total energy=max pe
1/2 m 14^2+m*g*29=mgh

solve for max height h

Answer 2

what are you referring to as far as m? i just started classes a few days ago im sorry, this is a cram semester so we are moving fast.

we havent discussed mass, i have studied average velocity, instantaneous velocity, acceleration, distance, and time

Answer 3

and we havent studied energy either

Answer 4

so here is what i tried and it seems logical, i set my initial velocity to 14m/s, final velocity to 0m/s, and gravity as well, gravity lol, 9.8m/s!

so doing so, am i able to square my velocity, and then divide it by 2*9.8?

or vi^2/2(g)?

and then i added the value i found to my original 29m, which gave me a total of 39m, which sounds like a good representation of the effect of gravity vs momentum in this instance.

Answer 5

to anyone who reads this post, my above formula did solve the problem correctly

Answer 6

Congrats!
What you did is valid, using strictly kinematics equations.
Basically, you got
H=29+14^2/(2g)=39

If you solved the energy equation above, where m (mass) cancels out on both sides of the equation, you will get H=(14^2/2+29g)/g=39
The two match exactly assuming no air resistance or friction.

This also underlines the advantage of informing us of what you have done, or which topic you're on when posting a question. In science, there are often multiple ways to find solutions.