Projectile motion while playing catch

In summary: A throw that just skims the ceiling is not prohibited either. The problem statement also does not prohibit "throws" where the ball is released at the ceiling. The resulting "throw" could be a bit strange, but the ball would go horizontally some distance before hitting the ground. So, the problem needs more specification.In summary, the conversation discusses a problem involving playing catch with a ball in a dormitory hallway. The problem involves finding the maximum horizontal distance the ball can travel without bouncing, given the initial speed of the ball and the height of the ceiling. The conversation includes equations and attempts at solving the problem, as well as corrections and clarifications on the equations used. It is noted that the problem statement may need more
  • #1
Daniel Luo
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Homework Statement



You are playing catch with a friend in the hallway of your dormitory. The distance from the floor to the ceiling is D, and you throw the ball with an initial speed v0=√(6gD). What is the maximum horizontal distance (in terms of D) that the ball can travel without bouncing? (Assume the ball is launched from the floor).

Homework Equations



Range = [v02*sin(2θ)] / g

y = y0 + xtan(θ)-1/2gx2/(v02*cos2(θ))

The Attempt at a Solution



I tried to use the equations and solve for x, without much luck. All my calculations ended up with a lot of unknows...
 
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  • #2
You are given v0, y0 and you know that y <= D. g is a constant. What other unknowns did you come up with? It's very hard to check your work if you don't supply it for review.

P.S.: What is x supposed to be? Shouldn't you have a variable for time somewhere?
 
  • #3
SteamKing said:
P.S.: What is x supposed to be? Shouldn't you have a variable for time somewhere?
Looks like Daniel has substituted for t using t = x / v0 cos(θ). But it has been done incorrectly in the v0 sin(θ) t term, resulting in v0 tan(θ) instead of x tan(θ).

Daniel, think about the fact that the ball must just avoid hitting the ceiling. What does that tell you about v0 sin(θ)?
 
  • #4
#Haruspex

Thanks for the correction. It is corrected now.
 
  • #5
Ok so I tried this:

The max. height is:

D = [(6gD)*sin(θ)] / (2g)

which simplifies to:

D = 3Dsin(θ)

Hence: sin(θ) = 1/3.

I tried using this for the horizontal distance:

R = [(6gD) * (2 cos(θ) sin(θ))] / g

I found cos(θ) by: √(12-(1/3)2) = (√8)/3

Next,

R = [ 6gD * 2 * (√8)/3 * 1/3*] / g = [(8√2)/3]D

But this answer is incorrect according to the answers which says R = (4√2)D.

Can you see what I've done wrong?
 
  • #6
Daniel Luo said:
Ok so I tried this:

The max. height is:

D = [(6gD)*sin(θ)] / (2g)

Your method is good, but check this equation.
 
  • #7
#CAF123

OF COURSE! It is the square of the y-component of the initial velocity! So it's sin squared theta. Thanks for pointing it out and letting me think my self :-).
 
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  • #8
Can you tell me how did you come up to this please ??
Daniel Luo said:
R = [(6gD) * (2 cos(θ) sin(θ))] / g
 
  • #9
Hoang said:
Can you tell me how did you come up to this please ??
It's quoted in post #1 as a standard equation (but in the sin(2θ) form. It is the range of a projectile at angle θ fired from ground level.
It is not hard to derive it from first principles. Just write the two usual horizontal and vertical distance at time t equations, set the vertical distance to zero and eliminate t. Discard the x=0 solution.
 
  • #10
Daniel Luo said:
The max. height is:

D = [(6gD)*sin(θ)] / (2g)
Is this the y- equation? And if it is, are we allowed to eliminate the vy0?
 
  • #11
Touuka said:
Is this the y- equation? And if it is, are we allowed to eliminate the vy0?
As noted in post #7, it is wrong.
 
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  • #12
haruspex said:
As noted in post #7, it is wrong.
Thank you!
 
  • #13
I can see no restriction that we cannot throw the ball totally horizontally. Does that game of catch implies/restricts the totally horizontal throw?
 
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  • #14
Adesh said:
I can see no restriction that we cannot throw the ball totally horizontally. Does that game of catch implies/restricts the totally horizontal throw?
Nice. "Without bouncing" allows rolling.
 
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FAQ: Projectile motion while playing catch

How does the angle of release affect the range of a projectile while playing catch?

The angle of release is a crucial factor in determining the range of a projectile while playing catch. The range will be maximum when the angle of release is 45 degrees. As the angle of release increases or decreases from 45 degrees, the range of the projectile will decrease.

What is the effect of initial velocity on the trajectory of a projectile while playing catch?

The initial velocity of a projectile while playing catch determines the height and distance it will travel. The higher the initial velocity, the higher the projectile will go and the farther it will travel. However, if the initial velocity is too high, the projectile may travel out of reach or cause injury.

How does air resistance affect the motion of a projectile while playing catch?

Air resistance or drag can significantly affect the motion of a projectile while playing catch. The force of air resistance acts opposite to the direction of motion, slowing down the projectile and causing it to land closer to the thrower. This effect is more significant for lighter and slower projectiles.

What role does gravity play in projectile motion while playing catch?

Gravity is the primary force that affects the trajectory of a projectile while playing catch. It pulls the projectile downwards, causing it to follow a curved path. The higher the initial velocity, the longer the projectile will remain in the air before gravity brings it down.

How do different factors, such as weight and size, impact the motion of a projectile while playing catch?

The weight and size of a projectile can affect its motion while playing catch in several ways. Heavier projectiles will have a greater inertia, making them more difficult to accelerate and change their direction. Smaller projectiles may be affected more by air resistance. The shape and aerodynamics of the projectile also play a role in its motion.

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