Ball be thrown to have maximum horizontal distance

[joemama69]

Homework Statement

A ball is thrown eastward across level ground. A wind blows horizontally to the east, and assume that the effect of the wind is to provide a constant force to the east, equal in magnitude to the weight of the ball. At what angle theta (to the horizxontal) should the ball be thrown to have maximum horizontal distance

Homework Equations

The Attempt at a Solution

is this correct

I believe the final vector for maximum distance should be v = i + j

The vector for the wind is w = mass(ball)i

So the vector of the ball being thrown pluss the wind vector, must equal the final vector v

mass(ball)i + xi + yj = i + j

x = 1-mass(ball), and y = 1

thus the thrown vector is (1-mass(ball))i + y

 

[rl.bhat]

You have to deal this problem as the projectile motion problem.
Only difference is the horizontal component of the velocity is not constant. The acceleration in this direction is also g.
Find the time of flight and range.
To find theta for maximum range, find the derivative of the range and equate it to zero.

 

[joemama69]

hmmmmm never thought of it that way... still stuck though

v_ox = vcos\theta + m_B, where m_B = mass of the ball

I do not understand how the horizontal acceleration is g, if i take that into account wouldn't that make the velocity v_ox = vcos\theta + m_Bg. could you please elaborate on that statement

 

[rl.bhat]

The time of flight = 2vsinθ/g.
Horizontal acceleration = mg.
So the range R = vcosθ*2vsinθ/g + 1/2*mg*(2sinθ/g)^2
Find dR/dθ and equate to zero to find θ for maximum range.

 

[joemama69]

could u elaborate on how you got the time of flight

x = \frac{2v^{2}cosQsinQ}{g} +\frac{2mv^{2}sin^{2}Q}{g}

x = \frac{2v^{2}}{g}[cosQsinQ + msin²Q]

dx/dQ = \frac{2v^{2}}{g}[2cos - 2QsinQ + 2msinQcosQ] + [cosQsinQ + msin²Q]

how do i set that to 0, i could do it if that m was not there. did i make a mistake

 

[rl.bhat]

CosqsinQ = 1/2*sin2Q
dx/dQ = 2v^2/g(cos2Q - msin2Q)
Now equate it to zero and find Q.

 

[rl.bhat]

The time of flight = 2vsinθ/g.
Horizontal acceleration = mg.
So the range R = vcosθ*2vsinθ/g - 1/2*mg*(2sinθ/g)^2
Find dR/dθ and equate to zero to find θ for maximum range.

dR/dθ = 2v^2/g( cos2θ - msin2θ )
Equate it to zero and find θ.

 

[joemama69]

ok

\frac{dx}{dQ} = \frac{2v^{2}}{g}[cosQ + msin2Q]

so i must get cosQ + msin2Q = 0


I still don't see how to solve for Q when the m is in the equation...


Also could you please elaberate on how you solved for t...

 

[rl.bhat]

R/dθ = 2v^2/g( cos2θ - msin2θ ) = 0
Or cos2θ = msin2
Or sin2θ/cos2θ = 1/m
so tan2θ = 1/m

 

[joemama69]

so

\theta = \frac{arctan(1/m))}{2}

The problem states there is a numberical answer

 

[rl.bhat]

The time of flight = 2vsinθ/g.
Horizontal acceleration = mg.
So the range R = vcosθ*2vsinθ/g + 1/2*mg*(2sinθ/g)^2
Find dR/dθ and equate to zero to find θ for maximum range.

Sorry. I have edit this post.
Horizontal acceleration is g, not mg. So the range is given by

So the range R = vcosθ*2vsinθ/g - 1/2*g*(2vsinθ/g)^2
=V²*sin2θ/g - 2v²sin²θ/g
= v²/g[sin2θ - 2sin²θ]
now find dR/dθ and equate it to zero.

 

[joemama69]

x = v₂/g[sin2Q + 2sin²Q]

dx = v₂/g[2cos2Q + 4sinQcosQ] = 0

Q = 67.5

 

[rl.bhat]

x = v₂/g[sin2Q + 2sin²Q]

dx = v₂/g[2cos2Q + 4sinQcosQ] = 0

Q = 67.5

It is not correct.
2cos2Q - 2sin2Q = 0, because the force of wind opposes the motion.

 

[joemama69]

The ball is thrown eastward... a wind blows horizontally to the east...

doesnt that mean that the ball and the wind are in the same direction, not opposing

also please explain your reasoning for find t, it is really buggin me

 

[rl.bhat]

The ball is thrown eastward... a wind blows horizontally to the east...

doesnt that mean that the ball and the wind are in the same direction, not opposing

also please explain your reasoning for find t, it is really buggin me

I am sorry. I didn't notice that.
While going up the particle experiences two accelerations. Vertical acceleration reduces the vertical component of the velocity. The horizontal acceleration has no effect on vy. So the time of will not change due to the horizontal acceleration.

 

[joemama69]

ok so could u explain the t now...

do you agree with my answer