Projectile Motion Problems Involving angles

In summary, an archer shooting an arrow parallel to the ground has a y-coordinate of 3.00 meters, a theta-coordinate of 3 degrees, and a vx-coordinate of 74.5 meters/second.
  • #1
jtredz518
3
0
There are two problems involving projectile motion and angles I am having problems with

Homework Statement



You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 61.0 away, making a 3.00 angle with the ground.

x = 61m
Theta = 3 degrees

Homework Equations


y = (1/2)a(t^2)
y = xtan(theta)
Vox = x/t

The Attempt at a Solution


I first used y = xtan(theta) and got y=(61)tan(3)=3.196m From there i used y=(1/2)a(t^2) to solve for t and got t=0.808s. from there i used Vox=(61)/(0.808) and got 74.5m/s. The webste claims this is wrong. Any help?


Problem 2:

Homework Statement


A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 6.00 above the ground. The ball lands 30.0 away. What is his pitching speed?

I solved that part and got V=21.7m/s then there is the second part of the question.

As you think about it, you’re not sure he threw the ball exactly horizontally. As you watch him throw, the pitches seem to vary from 5° below horizontal to 5° above horizontal. What is the range of speeds with which the ball might have left his hand? Enter the minimum and the maximum speed of the ball.

y=6m
x=30m
V=21.7m/s?
Theta=5 degree?

Homework Equations


y = Vyt + (1/2)a(t^2)
Vy=Vtan(theta)
Vox = x/t

The Attempt at a Solution


I honestly don't really know what to do for the second part. I tried to find the velocity for if he had thrown 5 degree above the parallel.
So i got Vy=Vtan(5)=2.4
Then i used Y=Vyt+(1/2)a(t^2)=2.4t+(-4.9)(t^2) and solved for t and got t=1.37s and then used V=x/t=30/1.37=21.9m/s

I don't know if that's right because i have yet to press submit because i can not for the life of me figure out what the velocity would be if he threw the ball 5 degrees below the parallel
 
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  • #2
I first used y = xtan(theta)
Th above statement is wrong. θ is the angle made by the tangent drawn at the point of impact arrow to the ground.
tanθ = vertical velocity/ horizontal velocity.
If t is the time of flight, x is the range of the projectile, find the vertical velocity and the horizontal velocities.
 
  • #3
How would I go about doing that?
 
  • #4
Okay i got the arrow one. can anyone help with the baseball one?
 
  • #5
Refer second thread below. ( Projectile motion, angle of projection problem.)
 

Related to Projectile Motion Problems Involving angles

1. What is projectile motion?

Projectile motion is the motion of an object through the air that is affected by gravity. It follows a curved path known as a parabola.

2. How is the angle of projection related to the range of a projectile?

The angle of projection is directly related to the range of a projectile. The range is maximum when the angle of projection is 45 degrees. As the angle increases or decreases from 45 degrees, the range decreases.

3. How do you calculate the maximum height of a projectile?

The maximum height of a projectile can be calculated using the formula h = (v2 * sin2θ) / 2g, where h is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

4. Can the angle of projection affect the time of flight of a projectile?

Yes, the angle of projection can affect the time of flight of a projectile. The time of flight is maximum when the angle of projection is 90 degrees, and decreases as the angle decreases from 90 degrees.

5. How do you solve for the components of a projectile's velocity at a given angle?

To solve for the components of a projectile's velocity at a given angle, you can use the formula vx = v * cosθ and vy = v * sinθ, where vx is the horizontal component of velocity, vy is the vertical component of velocity, v is the initial velocity, and θ is the angle of projection.

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