11. [-/4 Points] DETAILS MY NOTES SCALCET9 2.7.012.MI. If a rock is thrown upward on the planet Mars with a velocity of \( 15 \mathrm{~m} / \mathrm{s} \), its height above the ground (in meters) a \( t \) seconds is given by \( \mathrm{H}=15 \mathrm{t}-1.86 \mathrm{t}^{2} \). (a) Find the velocity (in \( \mathrm{m} / \mathrm{s} \) ) of the rock after 2 seconds. (b) Find the velocity (in \( \mathrm{m} / \mathrm{s} \) ) of the rock when \( t=a \). (c) When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.) ( t . \( \mathrm{m} / \mathrm{s} \) (d) With what velocity (in \( \mathrm{m} / \mathrm{s} \) ) will the rock hit the surface? (d \( \mathrm{m} / \mathrm{s} \)
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To dive into the rock's journey on Mars, let's first calculate its velocity. The velocity of the rock at any time \( t \) can be found by taking the derivative of the height function \( H \) with respect to \( t \): **(a)** The height function is \( H = 15t - 1.86t^2 \). The velocity function \( V(t) \) is \( V(t) = \frac{dH}{dt} = 15 - 3.72t \). Plugging in \( t = 2 \) seconds, we have: \[ V(2) = 15 - 3.72(2) = 15 - 7.44 = 7.56 \, \text{m/s} \] **(b)** For the general case at \( t = a \), the velocity is: \[ V(a) = 15 - 3.72a \] **(c)** To determine when the rock hits the surface, set \( H = 0 \): \[ 0 = 15t - 1.86t^2 \] Factoring gives us: \[ t(15 - 1.86t) = 0 \] So, \( t = 0 \) (launch time) or \( t = \frac{15}{1.86} \approx 8.1 \) seconds. **(d)** Finally, at \( t = 8.1 \) seconds, the velocity is: \[ V(8.1) = 15 - 3.72(8.1) \approx 15 - 30.12 = -15.12 \, \text{m/s} \] (The negative indicates it's moving downward.) So, in summary: - After 2 seconds, the velocity is **7.56 m/s**. - The velocity at \( t = a \) is **\( 15 - 3.72a \, \text{m/s}**. - The rock hits the surface at **8.1 seconds**. - The rock hits the surface with a velocity of **-15.12 m/s**.
