# Projectile motion while playing catch

## 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 alot of unknows...

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SteamKing
Staff Emeritus
Homework Helper
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?

haruspex
Homework Helper
Gold Member
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)?

#Haruspex

Thanks for the correction. It is corrected now.

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?

CAF123
Gold Member
Ok so I tried this:

The max. height is:

D = [(6gD)*sin(θ)] / (2g)
Your method is good, but check this equation.

#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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Can you tell me how did you come up to this please ??
R = [(6gD) * (2 cos(θ) sin(θ))] / g

haruspex
Homework Helper
Gold Member
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.

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?

haruspex
Homework Helper
Gold Member
Is this the y- equation? And if it is, are we allowed to eliminate the vy0?
As noted in post #7, it is wrong.

As noted in post #7, it is wrong.
Thank you!

I can see no restriction that we cannot throw the ball totally horizontally. Does that game of catch implies/restricts the totally horizontal throw?

jbriggs444