Projectile motion angle problem

theultimate6

A projectile is thrown from a point P at an angle θ above the horizontal . It moves in such a way
that its distance from P is always increasing from its launch until its fall back to the ground. Find ALL the possible values of θ with which the projectile could have
been thrown. You can ignore air resistance.

My attempt.

y=Vosinθ*t -0.5gt^2

x=Vocosθ*t

√(x^2 + y^2) = √((Vocosθ*t)^2+(Vosinθ*t-0.5gt^2)^2)

We have to take the derivative of the magnitude to time

dD/dt √((Vocosθ*t)^2+(Vosinθ*t-0.5gt^2)^2)

I want dD/dt > 0 so it is always positive

after derivation i get

t(g^2 * t^2 -3Vo*g*sin(θ)t +2Vo^2)>0

Let the discriminant be ∇

Now this polynomial here is positive when t>0 if ∇<0

∇=9Vo^2 * g^2 *sin(θ)^2 -8Vo^2 *g^2

∇<0 ∴ sin(θ)<√(8/9)

Solving for θ

θ<70.53

So i got the maximum value with is 70.53 degrees, what should i do next?

THERMO Spoken Here

Interesting...

Hi, You answer is "less than" some 70 degrees. Less than could be very many angles. The projectile starts at P, with elevation z @ P. At what elevation does it land? Just my thought about this before dinner...

ChipB

PHF Helper
Another way to approach this:

For a projectile launched at x = 0 the path the projectile takes is a parabola of the form y = Ax^2 + Bx, with the constants A and B governed by the initial velocity and angle of the shot. At any point along that arc the line from the launch point to the projectile has a slope of y/x, or Ax + B. The direction of travel of the projectile is given by dy/dx = 2Ax + B. Now, the projectileis moving away from the origin if the angle between the direction of travel and the line from the origing is greater than 90 degrees. If that angle is less than 90 degrees then the distance from the origin to the projectile is decreasing. So what we want o know is whether at any point along its path this angle is exactly 90 degrees, or perpendicuar. Remember that the sloped of two lines that intersect perpendicularly are the negative inverse of each other, so what we want to know is whether at any point does dy/dx = -x/y, or

2Ax + B = -1/(Ax +B)

I haven't taken it further than this, but perhaps ths will help.

ChipB

PHF Helper
Pursuing this further - the method I described above does indeed work well. You end up with a quadratic:

2A^2x^2 + 3ABx + (B^2+1) = 0

This has a a real solution only if (3AB)^2 - 8A^2(B^2+1) > 0. Note that we can divide through by and solve for B, yielding:

B^2 = 8

Now the last step is to determine how B relates to the initial conditions of velocity V_0 and angle of firing theta_0. From y(x) = Ax^2 + Bx, y'(x) = B, so at x=0 y'(0) = B, and this is the slope of the parabola at x=0. Hence B = tan (theta_0). So we have: tan(theta_0) = sqrt(8), from which theta_0 = 70.53 degrees. Hence any firing angle less than 70.53 degrees will result in a path where its distance from the firing point is always increasing.

ChipB

PHF Helper
Solving for θ

θ<70.53

So i got the maximum value with is 70.53 degrees, what should i do next?
You're done! The angle theta is the only variable that matters here; the initial velocity doesn't effect the outcome.

Aashwin

I neither get the question nor the solution. What actually was the question trying to elucidate and what answer was it expecting for??

ChipB

PHF Helper
I neither get the question nor the solution. What actually was the question trying to elucidate and what answer was it expecting for??
Depending on the angle with which you throw a ball it may either (a) continually move away from you throughout its entire flight (i.e. the distance from you to the ball is constantly increasing), or (b) if thrown at a steep enough angle it may initially move away as it gains altitude and then after it reaches its apex it may decrease its distance from you as it falls back to the ground. Consider the two trajectories in the attached figure - note that the blue arc has ever-increasing distance from where it is thrown (A < B < C < D < E) whereas the green arc has decreasing distances (A > B > C > D > E). So the problem is to figure out the launch angle that differentiates these two. Hope this helps.

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