QuestionJune 2, 2025

A stone is thrown with an initial velocity of 36ft/s from the edge of a bridge that is 49 ft above the ground. The height of this stone above the ground t seconds after it is thrown is f(t)=-16t^2+36t+49 If a second stone is thrown from the ground, then its height above the ground after t seconds is given by g(t)=-16t^2+v_(0)t where v_(0) is the initial velocity of the second stone Determine the value of v_(0) such that the two stones reach the same maximum height. When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its square will be zero. Therefore to find the maximum height of the object thrown from the bridge,use the square equation, square

A stone is thrown with an initial velocity of 36ft/s from the edge of a bridge that is 49 ft above the ground. The height of this stone above the ground t seconds after it is thrown is f(t)=-16t^2+36t+49 If a second stone is thrown from the ground, then its height above the ground after t seconds is given by g(t)=-16t^2+v_(0)t where v_(0) is the initial velocity of the second stone Determine the value of v_(0) such that the two stones reach the same maximum height. When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its square will be zero. Therefore to find the maximum height of the object thrown from the bridge,use the square equation, square
A stone is thrown with an initial velocity of 36ft/s from the edge of a bridge that is 49 ft above the ground. The height of
this stone above the ground t seconds after it is thrown is f(t)=-16t^2+36t+49 If a second stone is thrown from
the ground, then its height above the ground after t seconds is given by g(t)=-16t^2+v_(0)t where v_(0) is the initial velocity
of the second stone Determine the value of v_(0) such that the two stones reach the same maximum height.
When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its
square  will be zero. Therefore to find the maximum height of the object thrown from the bridge,use the
square  equation, square

Solution
4.5(189 votes)

Answer

\( v_0 \approx 66.56 \, \mathrm{ft/s} \) Explanation 1. Determine maximum height of first stone The velocity at maximum height is zero. Use the derivative f'(t) = -32t + 36. Set f'(t) = 0 to find t: \[ -32t + 36 = 0 \] \[ t = \frac{36}{32} = \frac{9}{8} \] 2. Calculate maximum height of first stone Substitute t = \frac{9}{8} into f(t): \[ f\left(\frac{9}{8}\right) = -16\left(\frac{9}{8}\right)^2 + 36\left(\frac{9}{8}\right) + 49 \] \[ = -16 \times \frac{81}{64} + \frac{324}{8} + 49 \] \[ = -20.25 + 40.5 + 49 \] \[ = 69.25 \, \text{ft} \] 3. Set up equation for second stone's maximum height The maximum height occurs when g'(t) = -32t + v_0 = 0. Solve for t: \[ t = \frac{v_0}{32} \] 4. Calculate maximum height of second stone Substitute t = \frac{v_0}{32} into g(t) and set equal to 69.25 ft: \[ g\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) \] \[ = -\frac{16v_0^2}{1024} + \frac{v_0^2}{32} \] \[ = \frac{v_0^2}{64} \] 5. Solve for \( v_0 \) Set \frac{v_0^2}{64} = 69.25: \[ v_0^2 = 69.25 \times 64 \] \[ v_0^2 = 4432 \] \[ v_0 = \sqrt{4432} \approx 66.56 \]

Explanation

1. Determine maximum height of first stone<br /> The velocity at maximum height is zero. Use the derivative $f'(t) = -32t + 36$. Set $f'(t) = 0$ to find $t$: <br />\[ -32t + 36 = 0 \]<br />\[ t = \frac{36}{32} = \frac{9}{8} \]<br /><br />2. Calculate maximum height of first stone<br /> Substitute $t = \frac{9}{8}$ into $f(t)$:<br />\[ f\left(\frac{9}{8}\right) = -16\left(\frac{9}{8}\right)^2 + 36\left(\frac{9}{8}\right) + 49 \]<br />\[ = -16 \times \frac{81}{64} + \frac{324}{8} + 49 \]<br />\[ = -20.25 + 40.5 + 49 \]<br />\[ = 69.25 \, \text{ft} \]<br /><br />3. Set up equation for second stone's maximum height<br /> The maximum height occurs when $g'(t) = -32t + v_0 = 0$. Solve for $t$: <br />\[ t = \frac{v_0}{32} \]<br /><br />4. Calculate maximum height of second stone<br /> Substitute $t = \frac{v_0}{32}$ into $g(t)$ and set equal to 69.25 ft:<br />\[ g\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) \]<br />\[ = -\frac{16v_0^2}{1024} + \frac{v_0^2}{32} \]<br />\[ = \frac{v_0^2}{64} \]<br /><br />5. Solve for \( v_0 \)<br /> Set $\frac{v_0^2}{64} = 69.25$:<br />\[ v_0^2 = 69.25 \times 64 \]<br />\[ v_0^2 = 4432 \]<br />\[ v_0 = \sqrt{4432} \approx 66.56 \]
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