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An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acceleration motion of the arrow, where v0 is the initial speed of the arrow, v is the speed of the arrow as it is moving up in the air, h is the height of the arrow above the ground, t is the time elapsed since the arrow was projected upward, and g is the acceleration due to gravity (9.8 m/s2).1. What is the maximum height from the ground the arrow will rise to the nearest meter?2. How long will it take for the arrow to reach its maximum height to the nearest tenth of a second?

Answer :

AMB000

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

[tex]v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}[/tex]

From them we can get [tex]v^2=v_0^2+2ad[/tex].

We have:

[tex]v-v_0=at\\t=\frac{v-v_0}{a}[/tex]

And substitute:

[tex]x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}[/tex]

We multiply both sides by 2a, and continue:

[tex]2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2[/tex]

Being d the displacement [tex]x-x_0[/tex], we have [tex]v^2=v_0^2+2ad[/tex]

For our exercise, we will write this as:

[tex]d=\frac{v^2-v_0^2}{2a}[/tex]

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

[tex]d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m[/tex]

For the time we use:

[tex]t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s[/tex]

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