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Mariulka [41]
3 years ago
13

A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met

ers in t sec.
a. Find the​ rock's velocity and acceleration at time t.
b. How long does it take the rock to reach its highest​ point?
c. How high does the rock​ go?
d. How long does it take the rock to reach half its maximum​ height?
e. How long is the rock​ a loft?
Mathematics
1 answer:
Verdich [7]3 years ago
5 0

Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

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Answer:

a)    tan (157.5) = \frac{1-cos 315}{sin315}

b)

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c)

      sin^{2} (157.5) = \frac{1-cos (315) }{2}

d)

  cos 330° = 1- 2 sin² (165°)

       

         

Step-by-step explanation:

<u><em>Step(i):-</em></u>

By using trigonometry formulas

a)

cos2∝  = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ =  2 cos² ∝/2

cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}

b)

cos2∝  = 1- 2 sin² ∝

cos∝  = 1- 2 sin² ∝/2

sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

<u><em>Step(i):-</em></u>

Given

              tan\alpha = \frac{sin\alpha }{cos\alpha }

          we know that trigonometry formulas

        sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

         1- cos∝ =  2 sin² ∝/2

      Given

         tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }

put ∝ = 315

      tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }

     multiply with ' 2 sin (∝/2) both numerator and denominator

        tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2}  }{2sin(\frac{315}{2} cos(\frac{315}{2}) }

Apply formulas

 sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

  1- cos∝ =  2 sin² ∝/2

now we get

 tan (157.5) = \frac{1-cos 315}{sin315}

       

b)

          sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 330° above formula

             sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}

            sin^{2} (165) = \frac{1-cos (330) }{2}

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c )

         sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 315° above formula

             sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}

            sin^{2} (157.5) = \frac{1-cos (315) }{2}

           

d)

     cos∝  = 1- 2 sin² ∝/2

   put      ∝ = 330°

       cos 330 = 1 - 2sin^{2} (\frac{330}{2} )

      cos 330° = 1- 2 sin² (165°)

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