How Do You Calculate the Initial Speed of a Baseball in a 2D Motion Problem?

[wwommack]

Homework Statement

A ball player hits a home run, and the baseball
just clears a wall 8.00 m high located 122.5 m
from home plate. The ball is hit at an angle
of 39.9◦ to the horizontal, and air resistance is
negligible. Assume the ball is hit at a height
of 1.2 m above the ground.
The acceleration of gravity is 9.81 m/s2 .
a) What is the initial speed of the ball?
Answer in units of m/s.

Homework Equations

y=yo+voy*t-1/2gt^2

y= yo+sinLaTeX Code: \\theta *t-1/2gt^2

1/2gt/yo= voy

R= Voy*t=VosinLaTeX Code: \\theta *t

t= LaTeX Code: \\sqrt{} (2y/g)^2

The Attempt at a Solution

I plugged in my known variables but couldn't get anything

Please help

 

[Furby]

According to your equations, it looks like you began in the right direction, but then steered off course a bit.

One thing to note is you subtracted (1/2)*g*t². Be careful that you don't put -9.8m/s² as g, since you prematurely subtracted it.

You began with
y=y₀+v_0y*t+(1/2)*a_y*t²

then you substituted v_0y with sin(39.9). It should be v_0y=v₀*sin(39.9).

I'm not sure what you did next. I would like to see how you derived (1/2)*g*t/y₀=v_0y.
I'm pretty sure it's nonphysical, though. When I solved it, I pretended the ball started at a height of 0 (y₀=0m) and traveled a height of 8m-1.2m (y=6.8m) so your equation would have a 0 in the denominator in my case.

I'm not certain what you intend R to represent. You're correct in stating v_0y*t=v₀*sin(39.9)*t, which is actually the flaw you made earlier in substitution, leaving v₀ out.

t=sqrt(((2y)/g)²)? Well first the square root and the square cancel out becoming t=(2y)/g. I'm not sure how you derived this, and I'm pretty sure it's nonphysical. EDIT: Ah, I think you tried to find the initial velocity by substituting time from your original equation INTO your original equation. Well, again I'm pretty sure that you didn't rearrange the equation correctly, and you need a second equation with the same unknowns to do this type of substitution. What I think you did was, say x-y=1, thus x=1+y, thus 1+y-y=1, which of course 1=1, which says nothing about x or y.

It looks like you understand that you'll have two unknowns (t and v₀). I think you went back to your original equation and rearranged it after having attempted substituting for v_0y. I would continue working with just substituting for your unknowns and leave rearranging your original equation at the end. You just need to create two equations with two unknowns. It appears you were going in the right direction, but made some serious algebra errors. Perhaps try starting with the x components first. That will be your second equation, and much simpler equation.

By the way if there is an easier way to solve this problem, I'm sorry I don't see it. Otherwise this is a long way to get the answer. =)

 

[tomcue]

!

Based on the given information, we can use the equations of motion to solve for the initial speed of the ball. First, we can use the equation y=yo+voy*t-1/2gt^2, where y is the height at any time t, yo is the initial height, voy is the initial velocity in the y direction, and g is the acceleration due to gravity. We know that the initial height is 1.2 m and the final height is 8 m, so we can plug in these values and solve for voy:

8 = 1.2 + voy*t - 4.905*t^2

Next, we can use the equation 1/2gt/yo= voy, which relates the initial velocity in the y direction to the time of flight and the initial height. We know that the initial height is 1.2 m and the acceleration due to gravity is 9.81 m/s^2, so we can plug these values in and solve for voy:

voy = 4.905*t

Finally, we can use the equation R= Voy*t=VosinLaTeX Code: \\theta *t, which relates the range of the ball (R) to the initial velocity (Voy), the time of flight (t), and the angle of projection (LaTeX Code: \\theta). We know that the range is 122.5 m and the angle of projection is 39.9 degrees, so we can plug these values in and solve for Voy:

122.5 = 4.905*t*cos(39.9)*t

Solving for t, we get t = 5.22 seconds. Plugging this back into our equation for voy, we get voy = 25.7 m/s. Therefore, the initial speed of the ball is 25.7 m/s.