30º above the horizontal with initial velocity problem

In summary, a basketball player throws a ball at an angle of 30º above the horizontal. If the ball has an initial velocity of 29.4 m/s, how far does it travel? The ball travels 11.025 ft when the time of flight is 3 seconds.
  • #1
Nexion21
4
0

Homework Statement



A basketball player throws a ball at an angle of 30º above the horizontal. If the ball in has an initial velocity of 29.4 m/s, how far does it travel?

Homework Equations



Vyf^2 = Vyi^2 + 2(Ay)(Dy) solve for Dy and get

[Vyf^2 - (Vyi^2)] / (2(Ay)) = Dy

The Attempt at a Solution



0 - (29.4sin30)^2 / (2(-9.8)) = 11.025

29.4sin30 = 14.7 and then 14.7^2 = 216.09 and then 0 - 216.09 is -216.09/(2(-9.8)) = 11.025
 
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  • #2
'how far does it travel' refers to a horizontal distance. First start by computing the time of flight.
 
  • #3
Okay so my new attempt is Vyf = Vyi + AyT

0 = 29.4 + -9.8t

-29.4 = -9.8t

3 = t

So my time is 3, now just don't know what equation to use
 
  • #4
Nexion21 said:
Okay so my new attempt is Vyf = Vyi + AyT

0 = 29.4 + -9.8t

-29.4 = -9.8t

3 = t

So my time is 3, now just don't know what equation to use
Check your value for [itex] v_{yi}. [/itex]
 
  • #5
Sorry that this is going to seem hopeless but I just started taking an online course... What exactly does Vyi stand for? I thought it was the initial velocity in the vertical direction but that must be wrong
 
  • #6
Nexion21 said:
Sorry that this is going to seem hopeless but I just started taking an online course... What exactly does Vyi stand for? I thought it was the initial velocity in the vertical direction but that must be wrong
You are correct, [itex] v_{yi} [/itex] is the vertical component of initial velocity. It is given by [itex] v_{yi} = v_i\,\sin\theta. [/itex]
 
  • #7
Ooh sinθ is with it. So vyi = 14.7, which would mean t = 1.5

After that I tried using the dx = vx * t equation but that didn't work. What is the formula to solve for horizontal distance?
 
  • #8
What you have found is the time to the apex of the trajectory. To find the time for the whole flight, either multiply your time by 2, (valid because of symmetry of flight and constant horizontal component of velocity) or redo the whole calculation using the condition, [tex] v_{yi}\hat{y} = -v_{yf}\hat{y} [/tex]
Edit: To clarify, the vertical component of final velocity (when the ball lands back to the ground) is equal in magnitude but opposite in direction to the vertical component of initial velocity.
 
Last edited:

1. What does it mean to have a 30º angle above the horizontal in a velocity problem?

This means that the initial velocity of the object is at a 30º angle above the horizontal line. This angle is measured from the ground up, with the horizontal line being parallel to the ground.

2. How is the initial velocity determined in a 30º above the horizontal velocity problem?

The initial velocity is determined by breaking down the given velocity into its horizontal and vertical components. In this case, the horizontal component would be the initial velocity and the vertical component would be the initial velocity multiplied by the sine of the angle (30º).

3. What is the relationship between the initial velocity and the angle in a velocity problem?

The initial velocity and the angle have a direct relationship. As the angle increases, the initial velocity also increases. This is because the initial velocity is determined by breaking down the given velocity into its horizontal and vertical components, with the vertical component being directly affected by the angle.

4. How does the angle affect the trajectory of the object in a 30º above the horizontal velocity problem?

The angle affects the trajectory of the object by determining the initial direction of the velocity. In this case, with a 30º angle above the horizontal, the object's trajectory will have a higher vertical component compared to a trajectory with a smaller angle or no angle at all.

5. How can the 30º above the horizontal velocity problem be solved?

The 30º above the horizontal velocity problem can be solved using the kinematic equations of motion. The initial velocity, angle, and other given information can be plugged into these equations to solve for the object's position, velocity, or time at any given point in the motion.

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