A juggler throws a ball from an initial height of 4 feet with an initial vertical velocity of 30 feet per second, the height of the ball can be modeled by h= -16t^2 +vt+s where t is the time in seconds the ball has been in the air, v is the initial vertical velocity, and s is the initial height. Write an equation that gives you the height of the feet of the ball as a function of time since it left the jugglers hand? Then calculate if the juggler misses the ball how many seconds does it take to hit the ground?
Question
Answer:
1. h = -16t² + 30t + 4
2. 2 s
Step-by-step explanation:
1. Write an equation that gives you the height of the feet of the ball as a function of time since it left the jugglers hand?
Since the height moved by the ball, h is given by
h = -16t² + vt + s
Since v = initial vertical velocity = 30 ft/s and s = initial height = 4 ft, then
substituting the values of the variables into the equation for h, we have
h = -16t² + vt + s
h = -16t² + 30t + 4
2. Then calculate if the juggler misses the ball how many seconds does it take to hit the ground?
The ball hits the ground when h = 0
So, h = 0 ⇒
-16t² + 30t + 4 = 0
dividing through by -2, we have
8t² – 15t – 2 = 0
Factorizing, we have
8t² – 16t + t – 2 = 0
8t(t – 2) + (t – 2) = 0
(8t + 1)(t – 2) = 0
8t + 1 = 0 or t – 2 = 0
8t = -1 or t = 2
t = -1/8 or t = 2
Since t cannot be negative, t = 2 s
So, the ball hits the ground after 2 s.