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3 years ago

1. When a cannonball is fired, the equation of its pathway can be modeled by h = -1612 + 128t.

Mathematics

1 answer:

sashaice [31]3 years ago

4 0

Answer:

The maximum height of the ball is 256m

Step-by-step explanation:

Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t

At maximum height, the velocity of the ball is zero.

velocity = dh/dt

velocity = -32t + 128

Since v = 0 at maximum height

0 = -32t+128

32t = 128

t = 128/32

t = 4seconds

The maximum height can be gotten by substituting t = 4 into the modelled equation.

h = -16t² + 128t

h = -16(4)²+128(4)

h = -16(16)+512

h = -256+512

h = 256m

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3 years ago

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inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

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We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

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