A quarterback throws a football from a height of 6 feet with an initial velocity of 20 ft per second the height h of the ball at the time T seconds can be represented by the equation 850 equals -16 squared + 20 t + 6 if the ball is caught at 2 feet find the range of the function
Answers
Answer:
range of function is 2ft to 12.24 ft
Step-by-step explanation:
A quarterback throws a football from a height of 6 feet with an initial velocity of 20 ft per second the height h of the ball at the time T seconds can be represented by the equation 850 equals -16 squared + 20 t + 6 if the ball is caught at 2 feet find the range of the function
height h = -16t² + 20 t + 6
V = u + at
max height is reached when v = 0
u = 20ft/s = 20 * 0.3 m/s = 6m/s
a = -g = -10 m/s²
0 = 6 + (-10)t
t = 6/10 sec
t = 0.6 sec
Max height = -16(0.6)² + 20(0.6) + 6
= -5.76 + 12 + 6
= 12.24 ft
Ball is caught at 2 ft so min height = 2ft
so range of function is 2ft to 12.24 ft
2 = - 16t² + 20t + 6
=> 16t² -20t - 4 = 0
=> t² - 5t - 1 = 0
t = (5 ± √(25+4))/2
t = (5 ± √29)/2
t = (5 ± 5.385)/2
t = 5.1925 sec
Domain of function = ( 0 to 5.1925 secs)