Question

16. The height, h, in feet of an object above the ground is given by h = -16t² +64t+190, t≥0, where t is the time in seconds. a) b) c) d) When will the object be 218 feet above the ground? When will it strike the ground? Will the object reach a height of 300 feet above the ground? Find the maximum height of the object and the time it will take.​

Answer 1

the maximum height is 254 feet.

Answer 2

a) 0.5 seconds and 3.5 seconds.

b) 5.98 seconds (2 d.p.)

c) No.

d) 254 feet at 2 seconds.

Explanation:

Given equation:

`h=-16t^2+64t+190, \quad t \geq 0`

where:

• h is the height (in feet).
• t is the time (in seconds).

Part a

To calculate when the object will be 218 feet above the ground, substitute h = 218 into the equation and solve for t:

`\begin{aligned}\implies -16t^2+64t+190 & = 218\\-16t^2+64t+190-218& = 0\\-16t^2+64t-28 & = 0\\-4(4t^2-16t+7) & = 0\\4t^2-16t+7 & = 0\\4t^2-14t-2t+7 &=0\\2t(2t-7)-1(2t-7)&=0\\(2t-1)(2t-7)&=0\\\implies 2t-1&=0\implies t=(1)/(2)\\\implies 2t-7&=0 \implies t=(7)/(2)\end{aligned}`

Therefore, the object will be 218 feet about the ground at 0.5 seconds and 3.5 seconds.

Part b

The object strikes the ground when h is zero. Therefore, substitute h = 0 into the equation and solve for t:

`\begin{aligned}\implies -16t^2+64t+190 & = 0\\-2(8t^2-32t-95) & = 0\\8t^2-32t-95 & = 0\end{aligned}`

Use the quadratic formula to solve for t:

`\implies t=(-b \pm √(b^2-4ac) )/(2a)`

`\implies t=(-(-32) \pm √((-32)^2-4(8)(-95)) )/(2(8))`

`\implies t=(32 \pm √(1024+3040) )/(16)`

`\implies t=(32 \pm √(4064) )/(16)`

`\implies t=(32 \pm √(16 \cdot 254) )/(16)`

`\implies t=(32 \pm √(16) √(254) )/(16)`

`\implies t=(32 \pm 4 √(254) )/(16)`

`\implies t=(8\pm √(254) )/(4)`

As t ≥ 0,

`\implies t=(8+ √(254) )/(4)\quad \sf only.`

`\implies t=5.98 \sf \; s \; (2 d.p.)`

Therefore, the object strikes the ground at 5.98 seconds (2 d.p.).

Part c

To find if the object will reach a height of 300 feet above the ground, substitute h = 300 into the equation and solve for t:

`\begin{aligned}\implies -16t^2+64t+190 & = 300\\-16t^2+64t+190-300 & =0\\-16t^2+64t-110 & =0\\-2(8t^2-32t+55) & =0\\8t^2-32t+55& =0\end{aligned}`

Use the quadratic formula to solve for t:

`\implies t=(-b \pm √(b^2-4ac) )/(2a)`

`\implies t=(-(-32) \pm √((-32)^2-4(8)(55)) )/(2(8))`

`\implies t=(32 \pm √(1024-1760) )/(16)`

`\implies t=(32 \pm √(-736) )/(16)`

`\implies t=(32 \pm √(16 \cdot -1 \cdot 46) )/(16)`

`\implies t=(32 \pm √(16) √(-1) √( 46))/(16)`

`\implies t=(32 \pm 4i√( 46) )/(16)`

`\implies t=(8\pm √( 46) \;i)/(4)`

Therefore, as t is a complex number, the object will not reach a height of 300 feet.

Part d

The maximum height the object can reach is the y-coordinate of the vertex.

Find the x-coordinate of the vertex and substitute this into the equation to find the maximum height.

`\textsf{$x$-coordinate of the vertex}: \quad x=-(b)/(2a)`

`\implies \textsf{$x$-coordinate of the vertex}=-(64)/(2(-16))=-(64)/(-32)=2`

Substitute t = 2 into the equation:

`\begin{aligned}t=2 \implies h(2)&=-16(2)^2+64(2)+190\\&=-16(4)+128+190\\&=-64+128+190\\&=64+190\\&=254\end{aligned}`

Therefore, the maximum height of the object is 254 feet.

It takes 2 seconds for the object to reach its maximum height.