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A stone thrown downward with an initial velocity of $44.1 \mathrm{~m} / \mathrm{sec}$ will travel a distance of $s$ meters, where $s(t)=4.9 t^{2}+44$.1t and $t$ is in seconds. If a stone is thrown downward at $44.1 \mathrm{~m} / \mathrm{sec}$ from a height of $441 \mathrm{~m}$, how long will it take the stone to hit the ground?

It will take $\square$ seconds to hit the ground. (Simplify your answer.)

Step 1 ：The problem is asking for the time it takes for the stone to hit the ground when thrown from a height of 441 meters. This means we need to find the time 't' when the distance 's' is equal to 441 meters. We can do this by setting the equation for 's' equal to 441 and solving for 't'.

Step 2 ：Let's set up the equation: \(s = 4.9t^2 + 44.1t\)

Step 3 ：Solving this equation gives two values for 't', -15 and 6. Since time cannot be negative, we discard -15.

Step 4 ：Therefore, the time it takes for the stone to hit the ground is 6 seconds.

Step 5 ：Final Answer: It will take \(\boxed{6}\) seconds to hit the ground.