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Andrew is an avid archer. He launches an arrow that takes a parabolic path.
The equation of the height of the arrow with respect to time is
y = -4.9x2 + 48x, where y is the height of the arrow in meters above
Andrew's bow and x is the time in seconds since Andrew shot the arrow.
Find how long it takes the arrow to come back to a height even with his bow
height.

Sagot :

Answer:

9.7959 sec

Step-by-step explanation:

For the arrow to reach the same height as the bow again, - 4.9x^2+48x=0, 48=4.9x, x=48/4.9=9.7959

The time arrow take to come back to a height even with his bow height is 9.79 seconds.

We have an equation of the height of the arrow with respect to time -[tex]y = -4.9x^{2} +48x[/tex] where y is the height of the arrow in meters above Andrew's bow and x is the time in seconds since Andrew shot the arrow.

We have to find out - how long it takes the arrow to come back to a height even with his bow height.

The motion of arrow in the above situation is an example of which type of motion?

It is an example of two - dimensional Projectile motion.

We have the function that depicts the variation of height of the arrow with respect to time given by -

[tex]y=-4.9x^{2} +48x[/tex]

To find the time taken by the arrow to come to a height even with his bow height, we should equate y = 0.

[tex]y=-4.9x^{2} +48x=0\\-4.9x(x-9.79)=0\\-4.9x=0\;\;\;and\;\;\;x-9.79=0\\x =0\;\;\;and\;\;\;x=9.79[/tex]

Time cannot be 0, hence the time arrow take to come back to a height even with his bow height is 9.79 seconds.

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https://brainly.com/question/13630358

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