I need to figure out what the velocity will be as it hits the basket 3m off the ground.

Question

If a basketball if shot to a distance of 27.6m, with an initial velocity of 17.1 m/s at an angle of 40 degrees to the horizontal from a height of 2 meters above the ground, what is its velocity as it hits the basket 3m off the ground?

Answer 1

This is a problem that could be very complicated, depending on how you approach it. This is the simplest way, assuming that you have to use kinematics methods and NOT energy methods.

The first step is to split the initial velocity into components, so that you have the y- and x-velocities.

v_y = 17.1sin(40) = 11 m/s v_x = 17.1cos(40) = 13.1

Since the x-velocity is constant in projectile motion, we only need to find the y-velocity at the end.

We know the acceleration and the change in height (1 meter), so the quickest equation to use is this one:

v² = 2*a*y + v_i²

Plugging in our known variables and solving for final y-velocity:

v^2 = 2(-10)(1) + 11^2 = -20 + 121 = 101

v = √101 ≅ -10.04 m/s (negative since the basketball is presumably on the downward path).

The last bit of this problem is just to solve for the magnitude and direction of the velocity, if necessary.

As a vector, v = <13.1, -10.04>. This would be √(13.1^2 + -10.04^2) = √(169.2+101) = 16.4 m/s

at a direction of arctan(-10.04/13.1) which is 37.5 degrees below the horizontal.

You might notice that the horizontal distance to the hoop isn't strictly necessary here, as the use of that particular kinematics equation eliminates the need to solve for the time. If you do have to solve for the time (or if you want to double check that this is past the top of the arc), you will need that value.