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Sagot :
hello
to solve this question, we proceed to apply the formula given
initial velocity v0 =
[tex]\begin{gathered} v_0=10\text{ m/s} \\ s=660m \end{gathered}[/tex]next we proceed to substitute the values into the equation
[tex]\begin{gathered} s=4.9t^2+v_0t \\ 600=4.9t^2+10t \\ 4.9t^2+10t-600=0 \end{gathered}[/tex]we'll proceed to solve this quadratic equation to find the time it took the ball to hit the ground
there are several methods to solve a quadratic equation and for the purpose of this session, i'll make use of quadratic formula
[tex]t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]in our equation, we have
[tex]\begin{gathered} a=4.9 \\ b=10 \\ c=-600 \end{gathered}[/tex]let's substitute the values into the equation
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t=\frac{-10\pm\sqrt[]{10^2-4\times(4.9\times-600)}}{2\times4.9} \\ t=\frac{-10\pm\sqrt[]{100+11760}}{9} \\ t=\frac{-10\pm\sqrt[]{11860}}{9} \\ t=\frac{-10\pm108.9}{9} \\ t=\frac{98.9}{9}\text{ or t}=\frac{-118.9}{9} \\ t=10.99\text{ ot t}=-13.21 \end{gathered}[/tex]but since time can't have a negative value, the answer is approximately 10.99.
the time taken for the object to hit the ground is approximately 10.99s
b.
how far will it fall in 3s
to find the distance the object travelled in 3s, let's substitute the value of time in the original equation
[tex]\begin{gathered} s=4.9t^2+v_0t \\ t=3 \\ v_0=10\text{ m/s} \\ s=4.9(3)^2+10\times3 \\ s=4.9\times9+30 \\ s=44.1+30 \\ s=74.1m \end{gathered}[/tex]in 3s, the ball travelled a distance of 74.1m
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