Another Two Dimensional motion question

In summary: Try cosine.cos35 = x/17than would you use the final speed formulasquare root Vx^2+Vy^2In summary, the second baseman throws the ball to the first baseman at a speed of 17.0 m/s and an angle of 35.0° above the horizontal. Using trigonometry, the horizontal and vertical components of the ball's initial velocity can be found. To find the horizontal component, use cosine with the angle of 35.0° and the initial speed of 17.0 m/s. The vertical component can be found using sine with the same angle and speed. The horizontal component remains constant throughout the throw, while the vertical component changes as the ball reaches the peak of its arc
  • #1
hurtingBrain
23
0

Homework Statement


A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is made with an initial speed of 17.0 m/s at an angle of 35.0° above the horizontal. (Neglect air resistance.)


What is the horizontal component of the ball's velocity just before it is caught?




The Attempt at a Solution


i got the time but i keep getting the velocity wrong.
 
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  • #2
You may be making this too complicated. You're given a vector at an angle of 35 degrees; try finding its components.
 
  • #3
what do you mean by the compents?
 
  • #4
The vectors in the horizontal and vertical directions that make up the vector you're given. The vector you're using has a magnitude of 17 m/s, at an angle of 35 degrees to the horizontal. Use trig functions to find the individual components.
 
  • #5
would you do this?
sin35 = x/17 and solve for x
 
  • #6
No, the sine helps you solve for the vertical component. Try cosine.
 
  • #7
cos35 = x/17
than would you use the final speed formula
square root Vx^2+Vy^2
 
  • #8
You're making it too complicated~

X is your horizontal component here. Solve 17cos35=x and you're done.
 
  • #9
really that's the velocity
 
  • #10
Yep. Your original vector is a velocity (speed in a direction), so therefore the component vectors are also velocities.
 
  • #11
i see, but i always thought that if you throw it x velocity in the beginning that it would be the same velocity at the end
and thanks :)
 
  • #12
But the ball's thrown at an angle. So it'll hit the first baseman's hit at the same speed that it left the second baseman's hand, yes. Assuming the second baseman's hand when the ball leaves it has the same elevation as the first baseman's mitt when he makes the catch. However, the component vectors are a different thing - since you're neglecting air resistance, the horizontal velocity remains the same throughout the throw, but the vertical velocity changes as the ball reaches the peak of its arc.

ANYWAY, USE TRIG FUNCTIONS TO FIND COMPONENTS
 
  • #13
lol. thanks for the help
 
  • #14
No problem :)
 
  • #15
ah, can you help me on this one too
A swimmer runs horizontally off a diving board with a speed of 3.60 m/s and hits the water a horizontal distance of 1.81 m from the end of the board.

How high above the water was the diving board?
 
  • #16
Alright, you know the swimmer's horizontal velocity is 3.6 m/s and he travels 1.81 m before hitting the water. You also know that acceleration due to gravity is 9.8 m/s, and his/her vertical velocity initially is 0.

Figure out how long the swimmer takes to travel 1.81 m at 3.6 m/s. That's also how long it takes to get from initial elevation to water level - you've now got t.

Then use the following formula:

x = x0 + v0t + .5at2
 
  • #17
would you use thise equation to find t
V^2 = Vo^2 + 2at
 
  • #18
opps.. i mean V^2 =Vo^2 + 2aX
 
  • #19
nope scratch that the first one
 
  • #20
No, you just need to use v = d/t. Use your horizontal information to calculate time.
 
  • #21
^ You need the final velocity to use that equation. Find the time to reach horizontal displacement, then use that time to find the vertical displacement.
 
  • #22
so would the final velocity be 3.60 m/s
 
  • #23
The horizontal velocity component would be 3.6 m/s, but the swimmer would also have a vertical velocity component due to gravitational acceleration. Use trig.
 
  • #24
JoshMP:

That's not needed for this problem. The swimmer travels a set horizontal distance at a set horizontal velocity; that can be used to find t.
 
  • #25
how would i use trg if i only have one side and no degree
 
  • #26
so use x = Xo + Voxt + .5at
to find t
by setting Vox 3.60ms
 
  • #27
No no no. JoshMP is incorrect. You can find t using your horizontal velocity and displacement.
 
  • #28
There is no horizontal acceleration. So Xf= Xi + Vot

Once you have t, plug it into Yf= Yi +.5(-g)t^2 and solve for Yf.
 
  • #29
ok, than in the x = Xo + Voxt + .5at^2
i know that a = -9.8
and t = .5 because i did V = t/D
is Xo = o and Vox = 0?
 
  • #30
"so would the final velocity be 3.60 m/s"

That was the question I was answering earlier when I said to use trig. You need to know the y-component to calculate the final velocity. I was not talking about finding t.
 
  • #31
v = d/t, you've got it the wrong way around.

And no, x = 0. But initial velocity is zero.
 
  • #32
joshMp how would you use trig if you only have one number and no degree?
 
  • #33
You have the x component, which was given to you. Find the y component using a kinematic equation. You now have the x component and the y component of the final velocity vector. Vf^2= Vx^2 + Vy^2
 
  • #34
JoshMP said:
"so would the final velocity be 3.60 m/s"

That was the question I was answering earlier when I said to use trig. You need to know the y-component to calculate the final velocity. I was not talking about finding t.

JoshMP said:
^ You need the final velocity to use that equation. Find the time to reach horizontal displacement, then use that time to find the vertical displacement.

My apologies, I thought the initial post (the one I've quoted second) was in response to v=d/t.
 
  • #35
i'm confused, so in Vf^2 = Vx^2 + Vy^2
and you said that final is 3.60 m/s than what is Vx?
 
<h2>1. What is two dimensional motion?</h2><p>Two dimensional motion refers to the movement of an object in two different directions simultaneously, typically represented on a coordinate plane.</p><h2>2. How is two dimensional motion different from one dimensional motion?</h2><p>One dimensional motion only involves movement in one direction, while two dimensional motion involves movement in two different directions at the same time.</p><h2>3. What are some examples of two dimensional motion?</h2><p>Some examples of two dimensional motion include a ball being thrown in the air, a car driving on a curved road, or a plane flying in the sky.</p><h2>4. What are the equations used to calculate two dimensional motion?</h2><p>The equations used to calculate two dimensional motion include the displacement formula, velocity formula, and acceleration formula for both the x and y directions.</p><h2>5. How is two dimensional motion used in real life?</h2><p>Two dimensional motion is used in many real life situations, such as calculating the trajectory of a projectile, predicting the movement of objects in sports, and analyzing the motion of planets and satellites in space.</p>

1. What is two dimensional motion?

Two dimensional motion refers to the movement of an object in two different directions simultaneously, typically represented on a coordinate plane.

2. How is two dimensional motion different from one dimensional motion?

One dimensional motion only involves movement in one direction, while two dimensional motion involves movement in two different directions at the same time.

3. What are some examples of two dimensional motion?

Some examples of two dimensional motion include a ball being thrown in the air, a car driving on a curved road, or a plane flying in the sky.

4. What are the equations used to calculate two dimensional motion?

The equations used to calculate two dimensional motion include the displacement formula, velocity formula, and acceleration formula for both the x and y directions.

5. How is two dimensional motion used in real life?

Two dimensional motion is used in many real life situations, such as calculating the trajectory of a projectile, predicting the movement of objects in sports, and analyzing the motion of planets and satellites in space.

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