PLEASEE Helppp need answer

Question

Let t represent the time in seconds a ball is in the air and h(t) represents the balls height in feet above the ground. Then it's height can be modeled by the equation: h(t)= -16t^2+50t+5

a) what is the maximum height?

b) when does it reach its maximum height? 

c) how long after its been thrown does it hot the ground?

d) how high is it after 1.5 seconds?

Answer 1

In his equation height is given by h(t) and t represents time. The shape of this equation is that of a downward facing parabola because the coefficient of the first term is negative. Therefore, the maximum height is found by determining the vertex of the parabola which is given by b/2a or 50/(16*2). Thus after, 1.5625 seconds the ball reaches it maximum height of 44 feet (plug 1.5625 into formula.) 

 

Notice that the formula has a constant of 5 in it meaning that the ball was throw up at a 5 foot starting position above the ground. Thus, for the ball to hit the ground h(t) = -5. So, what is the value of t that gives that value? Here we just use the quadratic formula and find that t = 3.31 seconds. That makes sense doesn't it? 2*1.5625 would get the ball back to the starting position ( 3.125 sec) and the ball needs another .185 sec to fall the last 5 feet. What speed is that in MPH? That would be 5/.185*(3600/5280) = 18.4 MPH. That is a lot slower than a 95 MPH fast ball.

 

To find the height after 1.5 second just plug in 1.5 for t.

Answer 2

a)

 

We can take the derivative of h(t) and set it equal to zero to find the maximum height.

 

h'(t) = 0

 

-32t + 50 = 0

 

-32t = -50

     t = 1.56

 

Plug this value into h(t).

 

h(t) = -16(1.56)² + 50(1.56) + 5

      = 44.06

 

The maximum height is 44.06

 

 

b) 

 

The ball reaches it maximum height at 1.56 seconds, as obtained in the previous part.

 

 

c)

 

When it hits the ground, h(t) = 0.

 

0 = -16t² + 50t + 5

 

Use the quadratic equation to solve for t.

 

t = (-50 ± √(2500 - 4(-80))) / -32

t = (-50 ± √(2820)) / -32

t = (-50 ± 53.10) / -32

 

t = (-50 + 53.10) / -32                and           t = (-50 - 53.10) / -32

 

t = -0.097                                 and            t = 3.22

 

 

The ball hits the ground in 3.22 seconds.

 

 

d)

 

Plug in  t=1.5   into h(t).

 

This part I let you complete on your own.