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# If a rock is thrown vertically upward from the surface of Mars with velocity 15 m/s, its height after $t$ seconds is $h = 15t - 1.86t^2.$(a) What is the velocity of the rock after 2 s?(b) What is the velocity of the rock when its height is 25 m on its way up? On its way down?

## a) $7.56 \mathrm{m} / \mathrm{s}$b) $v\left(t_{1}\right)=15-3.72 t_{1} \approx 6.24 \mathrm{m} / \mathrm{s}$ [upward] and $v\left(t_{2}\right)=15-3.72 t_{2} \approx-6.24 \mathrm{m} / \mathrm{s}$ [downward]

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here we have an equation that represents the height of Iraq. And we want to find the velocity at time two seconds. And so to find the velocity in general, we take the derivative of the height and the derivative of 15 T would be 15 on the derivative of negative 1.8 60 squared would be negative 3.7 to teach. Now that we have the velocity equation, weaken substitute a to in there for tea and we end up with 7.56 and the units for this one would be meters per second. And for part B, we want to find the velocity of the rock when the height is 25. So the first thing we need to do is find the time when the height is 25. So we're going to take our hide equation substitute 25 for the height and solve for T, Let's get everything over to one side. 1.8 60 squared minus 15 T plus 25 equals zero. And unfortunately, this is not factory ble. And so we're going to use the quadratic formula. T equals 15/3 0.72 plus or minus the square root of 15 squared minus four times 1.86 times 25 over 3.72 We're gonna let our calculator do some work for us, and we get some approximate times of 5.711 seconds and 2.354 seconds. So remember the goal was then to find the velocity at each of those times. The earlier time is going to be the time when it was on its way up. So on its way up, the velocity of 2.354 is approximately 6.24 meters per second and on its way down, that's going to be for the later time. So when it's on its way down, the velocity is V of 5.711 and that will be negative 6.24 meters per second. It makes sense that it was positive velocity on the way up and a negative velocity on the way down

Oregon State University

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