Respuesta :
ANSWER
5.08 seconds after being thrown
EXPLANATION
Given:
• The initial vertical velocity of the rock, u = 18 m/s
,• The initial height of the rock, y₀ = 37 m
Find:
• The time it takes the rock to be at 2 m from ground level, t
The vertical displacement of an object is given by,
[tex]y=y_o+ut-\frac{1}{2}gt^2[/tex]Where g is the acceleration of gravity, 9.8 m/s².
So, in this case, the height of the rock is given by the equation,
[tex]y=37+18t-4.9t^2[/tex]We have to find for what value of t, y = 2,
[tex]37+18t-4.9t^2=2[/tex]Subtract 2 from both sides,
[tex]\begin{gathered} 37-2+18t-4.9t^2=2-2 \\ \\ 35+18t-4.9t^2=0 \end{gathered}[/tex]To solve this equation, we can use the quadratic formula,
[tex]\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}[/tex]In this case, a = -4.9, b = 18, and c = 35,
[tex]t=\frac{-18\pm\sqrt{18^2-4\cdot(-4.9)\cdot35}}{2\cdot(-4.9)}=\frac{-18\pm\sqrt{324+686}}{-9.8}=\frac{-18\pm\sqrt{1010}}{-9.8}\approx\frac{-18\pm31.78}{-9.8}[/tex]The two possible solutions are,
[tex]\begin{gathered} t_1=\frac{-18-31.78}{-9.8}=\frac{-49.78}{-9.8}\approx5.08 \\ \\ t_2=\frac{-18+31.78}{-9.8}=\frac{13.78}{-9.8}\approx-1.41 \end{gathered}[/tex]Of these two results, t₂ is inconsistent with the problem. Remember that t represents time, so it cannot be negative.
Hence, we can conclude that the rock will be at 2 m from ground level 5.08 seconds after being thrown.
