New memeber. ball trajectory puzzle

[charlie'sthe1]

hi there, new member and this is a bit off a puzzle to me (this isn't homework its something I am trying to use for a little simulation game haha)

ok so the puzzle involves a ball being projected, i am wanting to find the minumum hight it can travell with the longest distance (does that make sence)

so i started of using suvats
u = 45 (45 sin theta in vertical, 45 cos theta in horizontal)
a (in vertical) = - 9.8
s = ?
t = ?
v = ?

so tried to do a bit of simutationas eqations with the vertical eqation. i reckond that at half the time the ball would be at its greatest, so i put in for t - t=0.5t int the equation s=0.5ut-4.9t^2
i then did a bit of differentation to find a when gradient = 0 and got t and sin theta. made a simulatanios EQ with my horizontal. and got theta = 81.7 DEG
i knew that wasnt right haha

could anybody think of an alternitive way or what i should do? is it solvable even?.. is the Q. understandable? haha (been up a long time)

 

[HallsofIvy]

Your equation is incorrect. The height (assuming that it is projected from sy= 0) is given by sy= uyt- 4.9t^2. There is no "0.5" in the first term. The distance traveled (assuming that it is projected from sx= 0) is given by sx= uxt.

With your values, that would be sy= 45cos(theta)- 4.9t^2 and sx= 45sin(theta)t.
If you are asking how far the projectile will go, set sy= 0 and solve for t. Then calculate that sx using that t. Of course, those will depend upon theta. You say "i reckond that at half the time the ball would be at its greatest" is correct if you mean half the time until the ball hits the ground (sy= 0).

If you are trying to calculate the angle that gives the greatest range, find sx depending on theta, then differentiate with respect to theta, set it equal to 0, and solve for theta.

 

[charlie'sthe1]

woo hoo it worked (it was 45 DEG... i shouldve know haha )