Isaac Physics Kinematics Question Help

The answer to the attached question is 60 degrees. However, I am getting -60 degrees as the answer. Could someone please explain where I went wrong? PFA my working.

The answer to the attached question is 60 degrees. However, I am getting -60 degrees as the answer. Could someone please explain where I went wrong? PFA my working.

Not really sure what youve done. Id probably model the motion perpendicular to the roof only, so its initial perpendicular distance from the roof is 10cos(30) and you want to find the angle which corresponds to reaching s=0 in a mimumum time. Projecting the motion onto this direction, you should get a quadratic in t which depends on 60-theta, and it should be a case of reasoning about that value (the trig value) to give the minimum time.

It seems that you have wrongly assumed that the horizontal and vertical displacements form "a right-angled triangle with the angle 30 deg". Hope you can understand.

Attached is the quadratic I am getting after doing this. However, I cannot think of any trig reasoning to minimise t. 😅 How to go about it then?

Probably the simplest one is to note that the object is decelerating in this direction, and the deceleration is constant so does not depend on the angle of projection. So the mimimum time will occur when the (projected) initial velocity has what property?

The question part probably doesnt need it, but you could note that assuming the initial velocity is large enough to reach the roof, then it will intersect twice (in general) and you want the first one (root) both of which are positive. Youre solving a quadratic like
x^2 - bx + c = 0
where b, c>0 and you can vary b up to some max value. The smallest root is given by
b - sqrt(b^2-4c) / 2
and to justify that it gets smaller as you increase b, simply take the derivative wrt b and make that argument. Increasing b to some max value is equivalent then to saying what about the trig term?

I’m really sorry for the late reply. Why does the initial velocity need to have a minimum value (a large enough value)? Even if it would be really small, would the time not adjust accordingly to reach the required displacement?

Its a question where either its a laborious maths proof about minimisng the value of the smallest roort or a relatively simple insight about relative motion So you had the projected suvat
-g/2 cos(30) t^2 + v cos(60-theta) t - 10 cos(30) = 0
Neither the constant decceleration (-g cos(30)) nor the required displacement 10 cos(30) depend on theta. Only the size of the initial velocity, vcos(60 - theta), does so
-v <= projected initial velocity <= v
as -1 <= cos() <= 1. If you had two different initial velocities u1 and u2 where u1 > u2, then their relative acceleration is zero (both moving with the same de/acceleration), so v1 > v2 throughout the entire trajectory and the first particle with initial velocity u1 will travel a given displacement, s, quicker than the second particle with initial velocity u2 as the velocity is always greater (by a constant amount). So we want to make vcos(60 - theta) as large as possible so cos()=1, so theta=60.

Proving that the minimum root of the quadratic decreases as cos increases is doable, but I cant imagine they wanted you to do this. The precalculus reasoning is simpler / more insightful than a calculus slog. Just a simple insight that acceleration and displacement are independent of theta, so the largest initial velocity will get there quicker, using a simple relative motion argument