Finding Initial Velocity and Angle in 2D Kinematics Problem

[CocoonOHorror]

Homework Statement

A softball is hit over a third baseman's head with some speed v-sub0 at an angle theta above the horizontal. Immediately after the ball is hit, the third baseman turns around and begins to run at a constant velocity V = 7.00m/s. He catches the ball t=2.00sec later at the same height at which it left the bat. The third baseman was originally standing L=18.0 m from the location at which the ball was hit. Find v-sub0 Use g=9.81 m/s^2 for the magnitude of the acceleration due to gravity. also find theta.

Homework Equations

x =( initial velocity times cos theta)* time
y = initial velocity times sin theta)* time - .5GT^2

The Attempt at a Solution

i don't understand how to solve an equation with 4 variables and i only have values for 2 of them. i have time = 2 seconds, and distance = 32 meters. how do i get an angle and an initial velocity?

 

[Doc Al]

i don't understand how to solve an equation with 4 variables and i only have values for 2 of them. i have time = 2 seconds, and distance = 32 meters. how do i get an angle and an initial velocity?

There are two equations and two unknowns. Hint: What's y, measured from the starting height?

 

[CocoonOHorror]

is it 9.8 meters?

 

[Doc Al]

is it 9.8 meters?

No. How does the initial height compare to the final height?

 

[CocoonOHorror]

it seems like you just use the dude running at 7 m/s for 2 seconds to get the total distance (18m + 14m) but i guess that's not the right way to go about it?i really wish my physics prof spoke english!

 

[CocoonOHorror]

No. How does the initial height compare to the final height?

they are the same.

 

[Doc Al]

it seems like you just use the dude running at 7 m/s for 2 seconds to get the total distance (18m + 14m) but i guess that's not the right way to go about it?

No, that's perfectly correct. That gives you the value for x you'll need in the first equation.

What about y?

 

[Doc Al]

they are the same.

Yes! So, if you measure from the starting point, what's y? (What's the change in height?)

 

[CocoonOHorror]

Yes! So, if you measure from the starting point, what's y? (What's the change in height?)

zero?

 

[Doc Al]

zero?

Yes, y = 0 when t = 2.

 

[CocoonOHorror]

so i have:
32=(initial velocity*cos theta)*time
and
0= (initial velocity*sin theta)*time - .5GT^2
right?
but now what?

 

[CocoonOHorror]

wait, i have 16 = initial velocity * cos theta
and 9.8 = initial velocity * sin theta (or is it 4.9/2 = initial Velocity * sin theta ?)

(i think)
now what?

 

[Doc Al]

Play around with those equations and see if you can isolate one of the variables. There are several ways to go. (Try division.)

 

[CocoonOHorror]

OK, maybe 16/cos theta = 9.8/sin theta?

im so lost...

 

[CocoonOHorror]

am i warm?

 

[CocoonOHorror]

is it an easy Trigonometric solution? I am in Trig/precalc algebra in one class and we just started the Trig half, so maybe I am not familiar with a function to equate Sin/Cos?

 

[Doc Al]

OK, maybe 16/cos theta = 9.8/sin theta?

im so lost...

You're not that lost. Multiply both sides by sin theta. (What other trig function appears?)

 

[CocoonOHorror]

does Tan theta = 9.8/16 ?
am i on the right track?

 

[CocoonOHorror]

holy mackeral! thanks Doc Al, i really appreciate the help!

 

[Doc Al]

You got it.