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
