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A ball is thrown vertically upward from the top of a building 192 feet tall with an initial velocity of 64 feet per second. The distance \( s \) (in feet) of the ball from the ground after t seconds is \( s(t)=192+64 t-16 t^{2} \). (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down? (a) After seconds the ball strikes the ground. (b)

Ask by Hanson Barker. in the United States
Feb 28,2025

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Answer

(a) The ball strikes the ground after 6 seconds. (b) The ball passes the top of the building on its way down after 4 seconds.

Solution

We are given the height function   s(t) = 192 + 64t – 16t². (a) The ball strikes the ground when s(t) = 0. Setting the equation equal to zero:   192 + 64t – 16t² = 0. Rearrange the equation:   –16t² + 64t + 192 = 0. Multiply both sides by –1:   16t² – 64t – 192 = 0. Divide the entire equation by 16:   t² – 4t – 12 = 0. Solve the quadratic using the quadratic formula, t = [4 ± √(4² – 4·1·(–12))] / 2:   t = [4 ± √(16 + 48)] / 2    = [4 ± √64] / 2    = [4 ± 8] / 2. This yields two solutions:   t = (4 + 8)/2 = 12/2 = 6  and  t = (4 – 8)/2 = (–4)/2 = –2. Since time cannot be negative, the ball strikes the ground after 6 seconds. (b) The ball is thrown from the top of the 192-foot building (t = 0, s = 192). It then rises, reaches a maximum height, and falls back down. On its way up and on its way down, the ball will pass the height of the building (192 feet). We already have s(t) = 192 when   192 + 64t – 16t² = 192. Subtract 192 from both sides:   64t – 16t² = 0. Factor out 16t:   16t(4 – t) = 0. Thus, t = 0 or t = 4. Since t = 0 is the starting point, the ball passes the top of the building on its way down at t = 4 seconds. Final Answers: (a) The ball strikes the ground after 6 seconds. (b) The ball passes the top of the building on its way down after 4 seconds.

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To find when the ball strikes the ground, we need to set \( s(t) = 0 \) and solve the equation: \[ 0 = 192 + 64t - 16t^2 \] Rearranging gives us: \[ 16t^2 - 64t - 192 = 0 \] Dividing through by 16: \[ t^2 - 4t - 12 = 0 \] Now, applying the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 1, b = -4, c = -12 \): \[ t = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \] This simplifies to: \[ t = \frac{4 \pm \sqrt{16 + 48}}{2} \] \[ t = \frac{4 \pm \sqrt{64}}{2} \] \[ t = \frac{4 \pm 8}{2} \] This gives us two solutions: 1. \( t = 6 \) seconds (when it hits the ground) 2. \( t = -2 \) seconds (discarded as time can't be negative) For part (b), to find when the ball passes the top of the building again on its way down, set \( s(t) = 192 \): \[ 192 = 192 + 64t - 16t^2 \] This simplifies to: \[ 0 = 64t - 16t^2 \] \[ 0 = 16t(4 - t) \] This gives us \( t = 0 \) (the initial throw) and \( t = 4 \) seconds (when it descends back past the same height). So the answers are: (a) 6 seconds, (b) 4 seconds. --- The concept of projectile motion dates back to Aristotle and Galileo, who studied the influences of gravity and initial velocities long before formulas like these were established. In fact, Galileo is credited with laying the groundwork for kinematics, which is vital in understanding objects moving in two dimensions! You can see projectile motion at work in real life every time you toss a basketball towards a hoop! The path it takes is a parabola, similar to the motion we discussed with the ball from the building. Understanding these principles can help in sports, engineering, and even video game physics to create more accurate simulations of motion!

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