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
to anyone who reads this post, my above formula did solve the problem correctly
total energy=max pe
1/2 m 14^2+m*g*29=mgh
solve for max height h
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
and we havent studied energy either
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.
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.
To solve this problem, we can use kinematic equations. Specifically, we can use the equation for vertical displacement:
Δy = v₀y * t + (1/2) * a * t²
Where:
Δy is the vertical displacement or change in height
v₀y is the initial vertical velocity
t is time
a is acceleration due to gravity (which is approximately -9.8 m/s² for objects near Earth's surface)
In this problem, we are given the following information:
v₀y (initial vertical velocity) = 14.0 m/s
Δy (vertical displacement) = 29.0 m
a (acceleration due to gravity) = -9.8 m/s²
We need to find the maximum height (which occurs when the vertical velocity becomes zero). At this point, the displacement is also 29.0 m above the street level.
So, we can set the final vertical velocity (vf) to zero and rearrange the equation to solve for the time (t):
0 = v₀y + a * t
Simplifying the equation, we have:
t = -v₀y / a
Now we can substitute the values we know into the equation:
t = -14.0 m/s / -9.8 m/s²
Solving for t, we get:
t ≈ 1.43 s
Now that we have the time it takes to reach the maximum height, we can use it to calculate the maximum height. Let's use the first equation again, but this time we'll solve it for Δy:
Δy = v₀y * t + (1/2) * a * t²
Substituting the known values:
Δy = 14.0 m/s * 1.43 s + (1/2) * -9.8 m/s² * (1.43 s)²
Simplifying the equation, we get:
Δy ≈ 10.0 m
Therefore, the maximum height that the baseball reaches is approximately 10.0 meters above its starting point on the street.