Help with basic trigonometric (Physics) problem

Question

I am re-learning basic Physics and I would like to know if I followed the correct steps, so I can continue doing more exercises. The problem says:

"A person kicks a ball from the surface of a playing field in the angle of 55º compared to the horizontal level. The ball lands 50.0m from the starting position. If the game is played inside a building, how high must the roof be? (Air resistance ignored)"

I drawn a triangle whose base is 50 and the angle AB is 55º. We have two angles: 90º and 55º. Since the triangle has a total of 180, then the remaining angle is 180-90-55 = 35º.

Using cos 55º, I calculated the hypotenuse, which is 87.17. Having a side (a=50) and the hypotenuse (c=87.17) I calculate the remaining side (b, which is the height of the roof) using the Pythagorean theorem, and gives ~71.40 meters of height.

Is this correct?

Answer 1

The trajectory of the ball will be a parabola - not a triangle - say Y =ax^2 + bx + c. set the point where the ball is kicked off at the origin. the point where it lands is then (0,50) The dervative of this Y, Y', will be tan 55 at x=0 - thse conditions will give you a, b, and c. Now find the point where Y'=0, that's where the ball is highest. insert this value into Y, and get the height.

John, a, b, and c are constants that will determine what the parabola looks like. We know that 0 = a*0^2 + b*0 + c because we chose to have the parabola go through the origin. From this we already get c = 0. In the same way, we get a and b by ineserting Y(50)=0 because the ball is on the floor again fifty m away. Now Y'(x) = 2ax + b. And Y'(0) = b = tan 55...

Need more?

Answer 2

Referring to the diagram below:-

a, b and c will be determined by

1. The force applied to the ball (and we will in turn quote the initial velocity instead).
2. The initial angle (to the horizontal).
3. The gravitation pull as ball's deceleration (when going up) & acceleration (when down).
4. Distance traveled and is 50 in this case.
5. The highest that the ball can reach = maximum height of the room. This occurs when the upward velocity is (gradually reduced to) 0. It is f(-b/[2a]) 'vertically' above the starting point (if a, b c are known and f(x) = ax^2 + bx + x).