Question

A basketball tosses a ball to a teammate. The ball reaches a maximum height of 11 feet 3 seconds after it was released. If the ball is thrown from a height of 6 feet, write a quadratic equation that represents the height of the ball as a function of time

Answer 1

Answer:

h = - 16t² + 49.67t + 6

Step-by-step explanation:

The expression for the height of the basketball, h at time, t is given by

h - h₀ = u₀t - 1/2gt² where h₀ = initial height of ball = 6 feet, h = maximum height of ball = 11 feet, u₀ = initial vertical velocity of ball, t = time taken to reach maximum height = 3 s and g = acceleration due to gravity =9.8 m/s².

We need to find u₀ from u = u₀ -

So, h - h₀ = u₀t - 1/2gt²

h - h₀ + 1/2gt² = u₀t

u₀ = (h - h₀)/t + 1/2gt

substituting the values of the variables into the equation, we have

u₀ = (h - h₀)/t + 1/2gt

u₀ = (11 ft - 6 ft)/3s + 1/2 × 32 ft/s² × 3 s

u₀ = 5 ft/3s + 1/2 × 96 ft/s

u₀ = 1.67 ft/s + 48 ft/s

u₀ = 49.67 ft/s

Using h - h₀ = u₀t - 1/2gt² where h is the height of the basketball at time, t

h = u₀t - 1/2gt² + h₀

substituting the values of the variables into the equation, we have

h = u₀t - 1/2gt² + h₀

h = (49.67ft/s)t - 1/2(32 ft/s²)t² + 6 ft

h = 49.67t - 16t² + 6

h = - 16t² + 49.67t + 6