Question

A ball is thrown into the air from a height of 4 feet with an initial velocity of 63 ft/sec. The function that models this situation is h(t) = –16t2 + 63t + 4, where t is measured in seconds and h is the height in feet.

(a) What is the height of the ball after 2 seconds?
(b) What is the maximum height of the ball?
(c) When will the ball hit the ground?
(d) What domain makes sense for the function?

Answer 1

#1

• h(2)
• -16(2)²+63(2)+4
• -16(4)+126+4
• -64+130
• 66ft

#2

Convert to Vertex form of parabola y=a(x-h)²+k

• y=-16t²+63t+4

We need calculator now

• y=-16(t-1.969)²+66.02

On comparing

• Vertex=(h,k)=(1.969,66.02)

As a is negative so parabola is opening downwards hence vertex is maximum

• Max height=66.02ft

#3

When h(t) is 0 ball hits the ground

• -16t²+63t+4=0
• -16t²-t+64t+4=0

On solving we get t=-1/16 or 4

Take it positive

• t=4s

At 4s it reach the ground

#4

Domain is set of all t values (Time axis)

Here

ball starts from 0s and hits 0 again at 4s

So domain=[0,4]

Answer 2

(a) 66 ft

(b) 66.015625 feet

(c) 4 s

(d) 0 ≤ t ≤ 4

Explanation:

`h(t)=-16t^2+63t+4`

Part (a)

`t=2 \implies h(2)=-16(2)^2+63(2)+4=66`

Therefore, the height of the ball after 2 seconds is 66 feet

Part (b)

The maximum height will be the turning point of the parabola.

To find the turning point, differentiate the function:

`\implies h'(t)=-32t+63`

Set it to zero:

`\implies h'(t)=0`

`\implies-32t+63=0`

Solve for t:

`\implies 32t=63`

`\implies t=(63)/(32)=1.96875`

Input found value of t into the function and solve for h:

`\implies h(1.96875)=-16(1.96875)^2+63(1.96875)+4=66.015625`

Therefore, the maximum height of the ball is 66.015625 feet

Part (c)

The ball will hit the ground when h(t) = 0

`\implies -16t^2+63t+4=0`

`\implies 16t^2-63t-4=0`

`\implies 16t^2+t-64t-4=0`

`\implies t(16t+1)-4(16t+1)=0`

`\implies (t-4)(16t+1)=0`

`\implies t=4, t=-0.0625`

As time is positive, t = 4s only

Part (d)

As time is positive, and the ball hits the ground when t = 4 s, the domain should be restricted to: 0 ≤ t ≤ 4