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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?

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oladayoademosu

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.