What is the formula for Maximum range and Maximum height? Medals will be rewarded.

Question

paxpolaris · Answer

range & height of what? be specific

anonymous · Answer

in projectile motion

paxpolaris · Answer

as the projectile goes upwards it slows down ... til it's vertical velocity becomes 0. Then it falls back down

to find max height: you can use the formula involving the squares of the velocities

paxpolaris · Answer

$$\LARGE {v_y}^2={u_y}^2-2g\cdot y$$
where,
     $y$ is the vertical position or height you are solving for (assuming you throw the projectile from y=0).
     $u_y$ is the vertical component of the initial/ starting velocity.
     $v_y$ is the vertical velocity at the point you trying to calculate the height for. (at max height $v_y=0$)
     $g$ is acceleration due to gravity $(\approx9.8m/s^2$ or $32ft/s^2)$

$$\large \implies 0 = {u_y}^2 -2g \cdot y_\max$$$$\large \implies y_\max = {{u_y}^2 \over2g }$$

paxpolaris · Answer

to find max range: .... assuming  the initial velocity is constant, only the angle you throw the projectile can change.

the max range is when throwing at a 45 degree angle.

$$\Large x = {2u^2 \over g}\cdot \sin(\theta)\cdot \cos(\theta)$$$$\large \implies x_\max ={u^2 \over g} \ \dots \text{{at 45 degrees}}$$

where, 
  $x$ is range: the horizontal position where the projectile falls back on the ground
  $u$ is initial velocity|dw:1344579713847:dw|