If an object is thrown upward, dropped, or thrown downward and travels in a straight line subject only to gravity, the height \( h \) of the object above the ground (in feet) after \( t \) seconds is given by \( h=-16 t^{2}+v_{0} t+h_{0} \), where \( h_{0} \) is the height of the object when \( t=0 \) and \( v_{0} \) is the velocity of the object at time \( t=0 \). Note that an object that is dropped (rather than thrown downward) has initial velocity \( v_{0}=0 \) You are standing on a cliff that is 200 feet high. How long will it take a rock to reach the ground if (a) you drop it? (b) you throw it downward at an initial velocity of 40 feet per second? (c) How far does the rock fall in 2 seconds if you throw it downward with an initial velocity of 40 feet per second? (a) A rock will take \( \square \) seconds to reach the ground if you drop it from 200 feet. (Type an integer or decimal rounded to two decimal places as needed.) (b) A rock will take \( \square \) seconds to reach the ground if you throw it downward at an initial velocity of 40 feet per second from 200 feet. (Type an integer or decimal rounded to two decimal places as needed.) (c) If you throw the rock downward with an initial velocity of 40 feet per second, then it falls \( \square \) feet in 2 seconds. (ther

To solve this problem, we will use the given equation for the height of an object under the influence of gravity: \[ h = -16t^2 + v_0t + h_0 \] where: - \( h \) is the height of the object above the ground in feet, - \( t \) is the time in seconds, - \( v_0 \) is the initial velocity of the object in feet per second, - \( h_0 \) is the initial height of the object above the ground in feet. Given that the cliff is 200 feet high, we have \( h_0 = 200 \) feet. (a) If the rock is dropped, the initial velocity \( v_0 = 0 \) feet per second. We need to find the time \( t \) it takes for the rock to reach the ground. (b) If the rock is thrown downward at an initial velocity of 40 feet per second, we need to find the time \( t \) it takes for the rock to reach the ground. (c) If the rock is thrown downward with an initial velocity of 40 feet per second, we need to find the distance it falls in 2 seconds. Let's solve each part step by step. Solve the quadratic equation by following steps: - step0: Solve using square roots: \(-16t^{2}+0\times t+200=0\) - step1: Simplify: \(-16t^{2}+200=0\) - step2: Add or subtract both sides: \(-16t^{2}=-200\) - step3: Divide both sides: \(\frac{-16t^{2}}{-16}=\frac{-200}{-16}\) - step4: Divide the numbers: \(t^{2}=\frac{25}{2}\) - step5: Simplify the expression: \(t=\pm \sqrt{\frac{25}{2}}\) - step6: Simplify the expression: \(t=\pm \frac{5\sqrt{2}}{2}\) - step7: Separate into possible cases: \(\begin{align}&t=\frac{5\sqrt{2}}{2}\\&t=-\frac{5\sqrt{2}}{2}\end{align}\) - step8: Rewrite: \(t_{1}=-\frac{5\sqrt{2}}{2},t_{2}=\frac{5\sqrt{2}}{2}\) The time it takes for the rock to reach the ground if it is dropped is approximately 3.54 seconds. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(-16t^{2}+40t+200=0\) - step1: Factor the expression: \(-8\left(t-5\right)\left(2t+5\right)=0\) - step2: Divide the terms: \(\left(t-5\right)\left(2t+5\right)=0\) - step3: Separate into possible cases: \(\begin{align}&t-5=0\\&2t+5=0\end{align}\) - step4: Solve the equation: \(\begin{align}&t=5\\&t=-\frac{5}{2}\end{align}\) - step5: Rewrite: \(t_{1}=-\frac{5}{2},t_{2}=5\) The time it takes for the rock to reach the ground if it is thrown downward at an initial velocity of 40 feet per second is approximately 5 seconds. Calculate the value by following steps: - step0: Calculate: \(-16\times 2^{2}+40\times 2+200\) - step1: Multiply the terms: \(-2^{6}+40\times 2+200\) - step2: Multiply the numbers: \(-2^{6}+80+200\) - step3: Evaluate the power: \(-64+80+200\) - step4: Add the numbers: \(216\) The distance the rock falls in 2 seconds if it is thrown downward with an initial velocity of 40 feet per second is 216 feet.