Question

suppose a stone is thrown vertically upward from the edge of a cliff on earth with an initial velocity of 39.2 m/s from a height of 161.7 m above the ground. the height (in meters) of the stone above the ground t seconds after it is thrown is s(t)= - 4.9t² + 39.2t + 161.7. the objects velocity after t seconds is v(t)= - 9.8t + 39.2 meters per second. a. determine the velocity v of the stone after t seconds. b. when does the stone reach its highest point? c. what is the height of the stone at the highest point? d. when does the stone strike the ground? e. with what velocity does the stone strike the ground? f. on what intervals is the speed increasing? a. the objects velocity after t seconds is v(t)= - 9.8t + 39.2 meters per second. b. it takes the object 4 seconds to reach its highest point. c. the object reaches a maximum height of 240.1 meters. d. the object strikes the ground after 11 seconds. e. the object strikes the ground with the velocity - 68.6 meters per second. f. the speed is increasing on the interval(s) (simplify your answer. type your answer in interval notation. use integers or decimals for any numbers in the expression. use a comma to separate answers as needed.)

Explanation:

1. a. Determine the velocity \(v\) of the stone after \(t\) seconds:

• The velocity - function \(v(t)\) is given by the derivative of the height - function \(s(t)\). Given \(s(t)=-4.9t^{2}+39.2t + 161.7\).
• Using the power rule \(\frac{d}{dt}(at^{n})=nat^{n - 1}\), we have \(v(t)=\frac{d}{dt}s(t)=-9.8t + 39.2\).

1. b. When does the stone reach its highest point?

• At the highest - point, the velocity \(v(t)=0\).
• Set \(v(t)=-9.8t + 39.2 = 0\).
• Solve for \(t\):
• Add \(9.8t\) to both sides: \(39.2 = 9.8t\).
• Then \(t=\frac{39.2}{9.8}=4\) seconds.

1. c. The object reaches a maximum height of:

• Substitute \(t = 4\) into the height - function \(s(t)=-4.9t^{2}+39.2t + 161.7\).
• \(s(4)=-4.9\times4^{2}+39.2\times4 + 161.7\).
• First, calculate \(-4.9\times4^{2}=-4.9\times16=-78.4\).
• Then, calculate \(39.2\times4 = 156.8\).
• \(s(4)=-78.4 + 156.8+161.7=240.1\) meters.

1. d. The object strikes the ground after:

• Set \(s(t)=-4.9t^{2}+39.2t + 161.7 = 0\).
• Divide the entire equation by \(-4.9\) to get \(t^{2}-8t - 33 = 0\).
• Factor the quadratic equation: \((t - 11)(t+3)=0\).
• Set each factor equal to zero: \(t - 11 = 0\) or \(t + 3 = 0\).
• Since time \(t\geq0\), we take \(t = 11\) seconds.

1. e. The object strikes the ground with the velocity:

• Substitute \(t = 11\) into the velocity - function \(v(t)=-9.8t + 39.2\).
• \(v(11)=-9.8\times11 + 39.2=-107.8+39.2=-68.6\) meters per second.

1. f. The speed is increasing on the interval:

• The acceleration \(a(t)=v^\prime(t)=-9.8\) (constant acceleration due to gravity). The speed \(|v(t)|\) is increasing when the velocity and acceleration have the same sign.
• The velocity \(v(t)=-9.8t + 39.2\). The acceleration \(a=-9.8\lt0\).
• The velocity \(v(t)\) is negative when \(-9.8t + 39.2\lt0\), which gives \(t\gt4\). So the speed is increasing on the interval \((4,11)\).

Step1: Find the velocity function

Differentiate \(s(t)=-4.9t^{2}+39.2t + 161.7\) using power - rule to get \(v(t)=-9.8t + 39.2\).

Step2: Find the time at the highest point

Set \(v(t)=0\) and solve for \(t\): \(-9.8t + 39.2 = 0\), so \(t = 4\) seconds.

Step3: Find the maximum height

Substitute \(t = 4\) into \(s(t)\): \(s(4)=-4.9\times4^{2}+39.2\times4 + 161.7=240.1\) meters.

Step4: Find the time when the object hits the ground

Set \(s(t)=0\), divide by \(-4.9\) to get \(t^{2}-8t - 33 = 0\), factor as \((t - 11)(t + 3)=0\), and take \(t = 11\) seconds (since \(t\geq0\)).

Step5: Find the velocity when the object hits the ground

Substitute \(t = 11\) into \(v(t)\): \(v(11)=-9.8\times11+39.2=-68.6\) m/s.

Step6: Find the interval where speed is increasing

Since \(a=-9.8\lt0\), find when \(v(t)\lt0\). Solving \(-9.8t + 39.2\lt0\) gives \(t\gt4\), and since the object hits the ground at \(t = 11\), the interval is \((4,11)\).

Answer:

a. \(v(t)=-9.8t + 39.2\)
b. \(4\)
c. \(240.1\)
d. \(11\)
e. \(-68.6\)
f. \((4,11)\)