Projectile Motion

Question

A basketball player 2 m tall wants to make a basket from a distance of 12 m. The hoop is at a height of 3.05 m. If he shoots the ball (from a height of 2 m) at a 29.1 ◦ angle, at what initial speed must he shoot the basketball so that it goes through the hoop without striking the backboard? The acceleration due to gravity is 9.8 m/s 2 . Neglect air friction. Answer in units of m/s.

Answer 1

In this projectile motion problem, we are given a handful of kinematic values which we can organize as variables in 2D. The lowest number of then significant figures provided is 1, but I will present additional precision the result. The x-y coordinates will extend positively rightward and upward.

Initial x-position: x₀ = 0 m (we can set this as our point of reference)

Final x-position: x = 12 m

Initial y-position: y₀ = 2 m

Final y-position: y = 3.05 m

Initial velocity direction θ = 29.1 deg

y-acceleration a_y = -9.8 m/s²

x-acceleration a_x = 0 m/s² (none implied)

We can use 'v₀' as the magnitude of the initial velocity vector. This velocity can be decomposed into x- and y-components using right-triangle trigonometry: since the initial velocity direction is the angle (θ) above the horizontal, the component in the horizontal (x-) direction is proportional to the cosine of the angle, and the vertical (y-) component is proportional to the sine of the angle.

v_0x = v₀*cos(θ)

v_0y = v₀*sin(θ)

Since there is no acceleration in the x-direction, we can relate this initial velocity to the time elapsed given the horizontal distance traveled at a constant speed:

x - x₀ = x = v_x*t = v₀*cos(θ)*t -> t = x/(v₀*cos(θ))

This is a little cumbersome but approaching the end, as we can substitute this into the time-dependent kinematic equation of the y-direction, as we have the remaining variables for this statement and can isolate and solve for v₀:

y = y₀ + v_θ0y*t + 1/2 a_y*t²

y - y₀ = v₀*sin(θ)*(x/[v₀*cos(θ)]) + 1/2 a_y*(x/[v₀*cos(θ)])²

(y - y₀) = [tan(θ)*x] + 1/2 a_y*x²/(v₀²*cos²(θ))

v₀²*cos²(θ)*(y - y₀ - [tan(θ)*x]) = 1/2 a_y*x²

v₀² = 1/2 a_y*x² / (cos²(θ)*(y - y₀ - [tan(θ)*x]))

v₀ = √{ 1/2 a_y*x² / (cos²(θ)*(y - y₀ - [tan(θ)*x])) }

Plugging in the values we defined above, we should get 12.8 m/s as the magnitude of the initial velocity, or the initial speed, v₀.