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Part 1 - ApplicationPhysics shows that when an object is thrown upward with an initial velocity of v then its approximate height is given by this quadratic function.s = − 4. 9t^2+ vot + hA ball is thrown upward from the top of a building 15 meters high, at a velocity of 2.8m/sec. Answer thefollowing questions. Be sure to show and explain all work.a) Find the maximum height and the time in which it was attained (round to 3 decimal places)b) When does it reach the ground? (round to 3 decimal places)c) Sketch a graph to illustrate your answers.

Sagot :

Since the ball is thrown upward from a building 15 meters high with a speed of 2.8m/s, replace h=15 and v=2.8 into the equation:

[tex]s=-4.9t^2+2.8t+15[/tex]

a)

The vertex of the parabola has the coordinates of the maximum height and the time it takes for the object to reach the maximum height. Since the graph of a parabola is symmetrical about the vertical line that passes through the vertex, we can simply find the time that it takes for the object to go back to the height of 15 meters, which is twice the time that it takes for it to reach its maximum height.

Then, replace s=15 and solve for t:

[tex]\begin{gathered} 15=-4.9t^2+2.8t+15 \\ \Rightarrow0=-4.9t^2+2.8t \\ \Rightarrow0=t(-4.9t+2.8) \end{gathered}[/tex]

Then, the times at which the object is at a height of 15 meters are given by:

[tex]\begin{gathered} t_1=0 \\ \\ -4.9t_2+2.8=0 \\ \Rightarrow t_2=\frac{2.8}{4.9} \\ \Rightarrow t_2=0.57143\ldots \end{gathered}[/tex]

The time at which the object reaches the maximum height is:

[tex]t=\frac{t_1+t_2}{2}=\frac{0+0.57143\ldots}{2}=0.2857\ldots[/tex]

Replace t=0.2857... into the equation to find the maximum height:

[tex]\begin{gathered} s=-4.9(0.2857)^2+2.8(0.2857\ldots)+15 \\ =15.4 \end{gathered}[/tex]

Then, the maximum height is 15.4 meters and it takes 0.286 seconds to reach it.

b)

To find the time that it takes for the object to reach the ground, replace s=0 and solve for t:

[tex]\begin{gathered} 0=-4.9t^2+2.8t+15 \\ \Rightarrow t=\frac{-2.8\pm\sqrt[]{2.8^2-4(-4.9)(15)}}{2(-4.9)} \\ \\ \therefore t_1=\frac{2+\sqrt[]{154}}{7}\approx2.059 \\ \therefore t_2=\frac{2-\sqrt[]{154}}{7}\approx-1.487\ldots \end{gathered}[/tex]

Since the negative time can be intepreted as an instant before the ball is thrown, the only meaningful result is t=2.06. Then, the ball reaches the ground after 2.06 seconds.

c)

Use the fact that the graph passes through the points (0,15), (0.286,15.4) and (2.06,0) to sketch it:

View image RashedT585598
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