An object is thrown upward at a speed of 191 feet per second by a machine from a height of 14 feet off the ground. The height of the object after [tex]$t$[/tex] seconds can be found using the equation [tex]$h(t)=-16 t^2+191 t+14$[/tex]. Give all numerical answers to 2 decimal places.
When will the height be 327 feet? At [ ] seconds
When will the object reach the ground? [ ] seconds
When will the object reach its maximum height? [ ] seconds
What is its maximum height? [ ] feet
Answer (1)
The height of 327 feet is reached at t = 1.96 and t = 9.98 seconds.
The object reaches the ground at t = 12.01 seconds.
The object reaches its maximum height at t = 5.97 seconds.
The maximum height is 584.02 feet.
Explanation
Problem Analysis We are given the height function h ( t ) = − 16 t 2 + 191 t + 14 , where h ( t ) is the height of the object in feet after t seconds. We need to find the time when the height is 327 feet, when the object reaches the ground, when the object reaches its maximum height, and what its maximum height is.
Time at 327 feet To find the time when the height is 327 feet, we set h ( t ) = 327 and solve for t . This gives us the equation − 16 t 2 + 191 t + 14 = 327 . Rearranging the equation, we get − 16 t 2 + 191 t − 313 = 0 . Using the quadratic formula, we find the two possible times to be approximately t = 1.96 seconds and t = 9.98 seconds.
Time at Ground Level To find when the object reaches the ground, we set h ( t ) = 0 and solve for t . This gives us the equation − 16 t 2 + 191 t + 14 = 0 . Using the quadratic formula, we find the positive time to be approximately t = 12.01 seconds.
Time at Maximum Height To find the time when the object reaches its maximum height, we find the vertex of the parabola represented by h ( t ) . The t -coordinate of the vertex is given by t = − b / ( 2 a ) , where a = − 16 and b = 191 . Thus, t = − 191/ ( 2 ∗ − 16 ) = 5.96875 . Rounding to two decimal places, we get t = 5.97 seconds.
Maximum Height To find the maximum height, we substitute the t -value obtained in the previous step into the height function: h ( 5.96875 ) = − 16 ( 5.96875 ) 2 + 191 ( 5.96875 ) + 14 = 584.015625 . Rounding to two decimal places, we get 584.02 feet.
Examples
Understanding projectile motion is crucial in fields like sports and engineering. For example, when designing a catapult or analyzing a baseball trajectory, we use similar equations to predict the range and maximum height of a projectile. By adjusting launch angles and initial velocities, engineers and athletes can optimize performance. This problem demonstrates the principles behind these calculations, showing how quadratic equations model real-world scenarios.
