The height of a ball t seconds after it is thrown upward from a height 32 ft and

Question

with initial velocity 48 ft per second is h(t) = -16t^2+48t+3

1. Verify that h(1)=h(2) 
2. According to Rolle s Theorem, what must be the velocity at some time in the interval (1,2)? Find that time. 
[Make your case; show work]

Answer 1

1)

The Height of the h(t) = -16t² + 48t + 3.

If t =1

h(1) = -16(1)² + 48(1)+3

h(1) = -16 +48 +3

h(1) = 35.

If t = 2

h(2) = -16(2)² + 48(2)+3

h(2) = -16*4 +48*2 +3

h(2) = -32 + 96 + 3

h(2) = 35.

Therefore h(1) = h(2).

Answer 2

2)

The Height of the h(t) = -16t² + 48t + 3.

Initial Velocity = 48 m/sec.

Height = 32 ft.

According to Rolle's theorem

Suppose f(x) is a function that satisfies all of the following.

1) f(x) is continuous on the closed interval [a,b].

2) f(x) is differentiable on the open interval (a,b).

3) f(a) = f(b)

Then f'(c) = 0 where a<c<b.

Now the h(t) = -16t² + 48t + 3.

h(t) is continuous in the interval [1,2]

h(t) is differentiable in the interval (1,2)

Now h(1) must be equal to h(2)

If t =1

h(1) = -16(1)² + 48(1)+3

h(1) = -16 +48 +3

h(1) = 35.

If t = 2

h(2) = -16(2)² + 48(2)+3

h(2) = -16*4 +48*2 +3

h(2) = -32 + 96 + 3

h(2) = 35.

Therefore h(1) = h(2).

Then h'(c) = 0

-32c + 48 = 0

c = 1.5

Therefore time = 1.5 sec.

Now we need to calculate Velocity at t = 1.5 sec.

We know that v = u + at.

a is the accerlation and can be found differentiating h'(t).

h'(t) = -32c + 48

Differntiate h'(t) with respect to t.

h''(t) = -32

a = -32 ft/sec²

Now v = u + at

V = 48 + (-32)(1.5)

v = 0 ft/sec.

Therefore Velocity at time = 1.5 sec is 0 m/sec.