Question
upstudy study bank question image url

A stone is thrown vertically into the air at an initial velocity of \( 73 \mathrm{ft} / \mathrm{s} \). On a different planet, the height s (in feet) of the stone above the ground after t seconds is \( \mathrm{s}=73 \mathrm{t}-5 \mathrm{t}^{2} \) and on Earth it is \( \mathrm{s}=73 \mathrm{t}-16 \mathrm{t}^{2} \). How much higher will the stone travel on the other planet than on Earth? When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its maximum heights of the two objects, use the

Ask by Flynn Hall. in the United States
Mar 12,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The stone travels approximately 183.22 feet higher on the other planet than on Earth.

Solution

We begin with the given height functions: - On the other planet: \[ s=73t-5t^2 \] - On Earth: \[ s=73t-16t^2 \] The maximum height occurs when the derivative of \( s \) with respect to \( t \) is zero. --- **Step 1. Differentiate the height functions** For the other planet, differentiate: \[ \frac{ds}{dt}=73-10t. \] Set the derivative equal to zero: \[ 73-10t=0 \quad\Longrightarrow\quad t=\frac{73}{10}. \] For Earth, differentiate: \[ \frac{ds}{dt}=73-32t. \] Set the derivative equal to zero: \[ 73-32t=0 \quad\Longrightarrow\quad t=\frac{73}{32}. \] --- **Step 2. Calculate the maximum heights** The maximum height is obtained by substituting the time calculated back into the corresponding height function. *Other planet's height:* \[ h_{\text{planet}}=73\left(\frac{73}{10}\right)-5\left(\frac{73}{10}\right)^2. \] Rather than compute step‐by‐step, we can notice that for any quadratic of the form \[ s=vt-at^2, \] the maximum height occurs at \[ t=\frac{v}{2a}, \] and the maximum height is given by \[ h_{\text{max}}=\frac{v^2}{4a}. \] Thus, \[ h_{\text{planet}}=\frac{73^2}{4\cdot 5}=\frac{5329}{20}. \] Numerically, \[ h_{\text{planet}}\approx \frac{5329}{20}\approx 266.45\text{ ft}. \] *Earth's height:* \[ h_{\text{earth}}=\frac{73^2}{4\cdot 16}=\frac{5329}{64}. \] Numerically, \[ h_{\text{earth}}\approx \frac{5329}{64}\approx 83.27\text{ ft}. \] --- **Step 3. Compute the difference** Subtract the two maximum heights: \[ \Delta h = h_{\text{planet}} - h_{\text{earth}} = \frac{5329}{20} - \frac{5329}{64}. \] To combine these, factor out \(5329\): \[ \Delta h = 5329\left(\frac{1}{20}-\frac{1}{64}\right). \] Find a common denominator: \[ \frac{1}{20}-\frac{1}{64}=\frac{64-20}{20\cdot 64}=\frac{44}{1280}. \] Thus, \[ \Delta h=5329\cdot\frac{44}{1280}=\frac{5329\cdot44}{1280}. \] Calculating the numerator: \[ 5329\cdot44=234476. \] Thus, \[ \Delta h=\frac{234476}{1280}\approx183.22\text{ ft}. \] --- The stone travels approximately \(183.22\) feet higher on the other planet than on Earth.

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

Extra Insights

To find how much higher the stone will travel on the other planet compared to Earth, we need to determine the maximum height reached in both scenarios. On the planet described by the equation \( s = 73t - 5t^2 \), the height is maximized by finding the vertex of the parabola represented by this quadratic equation. The time to reach the maximum height can be found using the formula \( t = -\frac{b}{2a} \). Substituting \( a = -5 \) and \( b = 73 \), we get \( t = \frac{73}{10} = 7.3 \) seconds. Plugging that value back into the height equation gives: \[ s = 73(7.3) - 5(7.3)^2 \] Calculating this will yield the maximum height on that planet. On Earth, we use the formula \( s = 73t - 16t^2 \), where using the same vertex formula gives \( t = \frac{73}{32} \approx 2.28125 \) seconds, and plugging that time back into the height equation will give the maximum height on Earth. The difference between these two heights will show how much higher the stone travels on the other planet compared to Earth. To make things more relatable, just picture launching a ball straight up at a park. On our planet (Earth), it might peak at a certain height before gravity brings it back down, but on another planet, where gravity acts weaker (represented by the smaller coefficient of \( t^2 \)), that same ball could soar even higher, giving an enthusiastic boost to your throwing arm! Speaking of jumping, isn’t it fascinating how different celestial bodies have varying gravitational pulls? For instance, on the Moon, your weight is about 1/6th of what it is on Earth, so if you threw that same stone, it would fly even higher and longer. Next time you daydream about cosmic adventures, remember that your throw would make you the star of the show!

Related Questions

Latest Calculus Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy