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Problems for Motion with Constant Acceleration in Two and Three Dimensions

Problem : A basketball player who is standing 15 feet away from a basketball hoop is trying to make a basket. If the height of the hoop is 10 feet, and the height at which the player shoots the ball is 6 feet, at what angle and with what speed should the player shoot the ball?

Figure %: Diagram of a basketball player shooting a ball with speed v and angle θ .

Using the general formula for projectile motion worked out in this section, we find that the position function for a ball which is shot from a height of 6ft, with initial velocity vector v 0 , is:

x(t) = (0, 0, - g)t 2 + v 0 t + (0, 0, 6)

where we are taking the origin of coordinates to be at the player's feet, and the positive z -direction points upwards. If we choose the x -direction to point in the direction of the hoop, we can ignore the y -direction and reduce this to a two-dimensional problem.

For the player to make his basket, the trajectory of the ball must pass through the point (15, 0, 10) (since the basket is located 15 ft away and is 10 ft high). Therefore, we need to find v 0 = (v x, 0, v z) such that x(t) = (15, 0, 10) at some time t (since there is no motion in the y -direction we know that v y must be zero). To do this, we write out the one-dimensional equations for the x - and z -directions:

(1) 15 = v x t , and (2) 10 = - gt 2 + v z t + 6

From (2) we find that the ball reaches a height of 10 ft at a time t such that:

gt 2 - v z t + 4 = 0

Using the quadratic formula, we find two solutions for t :

t 1 =  and t 2 = .

t 2 , which is the earlier time, corresponds to when the ball passes through a height of 10 feet on its way up. However, the basket must be made as the ball is falling back down. Therefore, t 1 is the time we are interested in.

It is not enough for the ball to have a height of 10 ft, however, in order for the basketball player to make the basket. At the same time it reaches that height (i.e. at time t 1 ) it must also have traveled 15 ft in the x - direction. We are now in a position to solve for v x in terms of v z . Using (1), and plugging in the time t 1 , we find that:

v x = =

So, any initial velocity vector of the form

v 0 = (, 0, v z)

will ensure that the player makes his basket.
Figure %: Several solutions to the problem of a basketball player shooting to make a basket.
In other words, for every possible value of v z (i.e. for every value of v z such that is a real number), there is a solution. You were probably already aware of this intuitively because by shooting a basketball at different angles and with different speeds you are able to make the same basket in a variety of ways!

Problem : Assume that the basketball player in the previous problem has a special condition which allows him only to shoot basketballs at a speed of 10 ft/s. How might someone go about finding at what angle must he shoot the ball in order to make the basket?

Using the solution to the previous problem,

v 0 = (, 0, v z)

we need only require that the magnitude of v 0 is 10 ft/s. This will enable us to solve for a single value of v z , which will in turn provide us with a unique value for v x . Following familiar rules from trigonometry, by taking arctanv z/v x we can then get the angle at which the player must shoot. The details of these particular calculations become quite ugly and are left as an exercise.