This is my catapult, the launching angle in its original form( sits on the ground) is less than 90°(although the arm hits the stopping bar at an 90° .However, i observed that the path of the table tennis ball is a parabola shape. That means the ball is launched at an angle and not 90°):

1)if i do not pull the arm, the arm will be at rest at certain angle because of the spring. Assume the angle to be ?. is the launching angle of the catapult in its original form same as ??

2) if i raise the front of the catapult by certain angle, say X degrees. My lauching angle in this case will be X+?, am i correct? if not, explain plz

When the arm hits the stopping bar, the arm makes a 90° relative to the ground

samkatietowers

It is possible the ball leaves the arm at a different angle than in theory because the ball has momemtum as it curves and the arm may not hold it completely perpendicular until the release point. If you had a slow motion camera, you might see the ball leave the arm at a different angle.

Changing the angle of the entire device changes other factors, such as the gravitational forces applied to the ball and mechanism, but in general, your assumption seems sound.

peace chandelier0207

Your question is not absolutely clear. Am I right in assuming that ‘the launching angle’ is the angle of elevation (i.e. above the horizontal) that the tangent to the trajectory of the ball makes at the point where it parts company with the cup of the catapult ? I will assume that this is what you mean.

Simplistically, if ‘When the arm hits the stopping bar, the arm makes a 90° relative to the ground’, is true, one would expect the launching angle to be zero (you say 90 deg – maybe this is measured with respect to the vertical,or is the angle of the catapult’s arm relative to the horizontal, rather than the angle of the velocity vector of the ball relative to the ground).

If the ball rises after it leaves the catapult then the angle of launch must be greater than zero. To explain this, it could be possible that the centrifugal force of the ball against the cup causes the ball to slip radially out of the cup before the arm becomes vertical. A better way of stating this might be to say that the cup must exert a radial inward force to keep the ball in a circular path, but that the shape of the cup and ball make it impossible for the required force to be exerted once the angle of the arm becomes a certain value above the horizontal, so the ball leaves the cup before the arm reaches the stopping bar, and the launch angle is then not zero.

Tipping the front of the catapult upward might not change this behaviour, unless, that is, the arm hits the stopping bar before the ball slips out of the cup, as described above. So,on the basis of this scenario, I would say that as X is increased, initially ? remains constant, but beyond a certain value of X, ? becomes equal to X. Note that in my discussion of this, both X and ? are measured relative to the horizontal.

Hope this helps….