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Sagot :
Answer:
speed = 7.88 m/s
Explanation:
First, we need to find the initial velocity of the stone, so we will use the following equation:
[tex]\Delta x=v_0t-\frac{1}{2}gt^2_{}[/tex]Where Δx is the distance, t is the time, v0 is the initial velocity, and g is the gravity. So, solving for v0, we get:
[tex]\begin{gathered} \Delta x+\frac{1}{2}gt^2=v_0t \\ \frac{\Delta x+\frac{1}{2}gt^2}{t}=v_0 \end{gathered}[/tex]Now, replacing the values, we get;
[tex]\begin{gathered} \frac{40.8\text{ m +}\frac{1}{2}(9.8m/s^2)(3.8s)^2}{3.8s}=v_0 \\ \frac{40.8\text{ m +}\frac{1}{2}(9.8m/s^2)(14.44s^2)}{3.8s}=v_0 \\ \frac{40.8\text{ m +70.756 m}^{}}{3.8s}=v_0 \\ \frac{111.556\text{ m}}{3.8\text{ s}}=v_0 \\ 29.36m/s=v_0 \end{gathered}[/tex]Then, the velocity of the stone just before it hits the water can be found using the following equation:
[tex]\begin{gathered} v_f=v_0-gt \\ v_f=29.36m/s-9.8m/s^2(3.8s) \\ v_f=29.36\text{ m/s - 37.24 m/s} \\ v_f=-7.88\text{ m/s} \end{gathered}[/tex]Therefore, the speed of the stone is 7.88 m/s
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