Finding Initial Velocity and Angle in 2D Kinematics Problem

In summary, to solve for the initial velocity and angle of a softball hit over a third baseman's head with a constant velocity and caught 2 seconds later, two equations can be used: x = (v0*cos(theta))*t and y = (v0*sin(theta))*t - 0.5*g*t^2. Using the known values of time (2 seconds) and distance (32 meters), the initial height can be determined as y = 0. From there, using basic trigonometry, the equation cos(theta) = 16/v0 and sin(theta) = 9.8/v0 can be obtained. By dividing these equations and manipulating them, the initial velocity and angle can be solved for.
  • #1
CocoonOHorror
25
0

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?
 
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  • #2
CocoonOHorror said:
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?
 
  • #3
is it 9.8 meters?
 
  • #4
CocoonOHorror said:
is it 9.8 meters?
No. How does the initial height compare to the final height?
 
  • #5
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!
 
  • #6
Doc Al said:
No. How does the initial height compare to the final height?

they are the same.
 
  • #7
CocoonOHorror said:
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?
 
  • #8
CocoonOHorror said:
they are the same.
Yes! So, if you measure from the starting point, what's y? (What's the change in height?)
 
  • #9
Doc Al said:
Yes! So, if you measure from the starting point, what's y? (What's the change in height?)

zero?
 
  • #10
CocoonOHorror said:
zero?
Yes, y = 0 when t = 2.
 
  • #11
so i have:
32=(initial velocity*cos theta)*time
and
0= (initial velocity*sin theta)*time - .5GT^2
right?
but now what?
 
  • #12
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?
 
  • #13
Play around with those equations and see if you can isolate one of the variables. There are several ways to go. (Try division.)
 
  • #14
OK, maybe 16/cos theta = 9.8/sin theta?

im so lost...
 
  • #15
am i warm?
 
  • #16
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?
 
  • #17
CocoonOHorror said:
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?)
 
  • #18
does Tan theta = 9.8/16 ?
am i on the right track?
 
  • #19
holy mackeral! thanks Doc Al, i really appreciate the help!
 
  • #20
You got it. :approve:
 

FAQ: Finding Initial Velocity and Angle in 2D Kinematics Problem

What is 2D kinematics?

2D kinematics is the study of motion in two dimensions, where an object's position, velocity, and acceleration are measured across both the x and y axes.

How is 2D kinematics different from 1D kinematics?

1D kinematics only considers motion along one axis, while 2D kinematics takes into account motion along both the x and y axes. This allows for a more accurate description of an object's motion.

What are some common equations used in 2D kinematics?

The most commonly used equations in 2D kinematics are the equations for position, velocity, and acceleration in both the x and y directions. These can be derived from the basic kinematic equations for 1D motion.

How do you solve a 2D kinematics problem?

To solve a 2D kinematics problem, you will need to break down the motion into its x and y components. Then, you can use the equations for position, velocity, and acceleration in each direction to solve for the unknown variable.

What are some real-world applications of 2D kinematics?

2D kinematics has many applications in fields such as engineering, physics, and sports. For example, it can be used to analyze the trajectory of a projectile, the motion of objects in a roller coaster, or the movement of athletes in different sports.

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