How do you figure out the velocitcy of a basketball when it's going through the air?

Question

it only has 2 do with the velocity

Answer 1

You need to know the initial speed that it was thrown at (v_i), the angle above the horizontal that it was thrown at (theta), and the time that has gone by. Then you split up the velocity into x and y components.

v_x = v_i*cos(theta)
v_y = v_i*sin(theta) - g*t

So the velocity at any given time will be:
v = sqrt(v_x^2 + v_y^2)

Answer 2

There are two components to the velocity of any projectile in a gravitational field: vertical, which varies with time; and horizontal, which is constant if you can ignore the effects of air resistance.

If you know the angle at which the basketball was thrown and its initial velocity you can accurately predict how its height and speed change with time, or if you have a high-speed camera you can compare two or more adjacent frames to get the same information.

The vertical velocity component is given by v * sine(angle) and the horizontal by v * cosine(angle). V(vert) is affected by the acceleration of gravity, 9.8 meters/second/second. If you multiply this acceleration 'g' by the time in seconds you can find the instantaneous vertical velocity v = g * t, which can be either positive or negative as the ball goes up or down.

The vertical distance traveled at any time 't' is given by d = 1/2 g t^2 (because velocity is the derivative of distance with respect to time (and acceleration is the derivative of velocity with respect to time)).

You now have enough information to solve any simple projectile problem.

Answer 3

Well, if you know exactly where the ball is, then there is no way you can know exactly what the velocity is. ;-)

Answer 4

v=u+gt

v= final velocity
u=intial velocity
g= gravity
t= time elapsed