Solve the problem.
-An object is thrown upward from the top of a 160-foot building with an initial velocity of 48 feet per second. How long after the object is projected upward will it strike the ground? Use the equation $$h = \frac { 1 } { 2 } g t ^ { 2 } + v _ { o } t + h _ { o }.$$
A)-2 sec
B)5 sec
C)160 sec
D)2 sec

Question

Quizplus · Accepted Answer

The equation given is a model for the height of an object thrown upward with gravity acting on it, where $h$ is the final height, $g$ is the acceleration due to gravity (in feet per second squared, $g = 32 \, \text{ft/s}^2$ downward), $v_o$ is the initial velocity, $t$ is the time in seconds, and $h_o$ is the initial height. To find when the object hits the ground, set $h = 0$ and solve for $t$:$$0 = -16t^2 + 48t + 160.$$This is a quadratic equation in the form of $at^2 + bt + c = 0$, where $a = -16$, $b = 48$, and $c = 160$. Solving this equation for $t$ gives the time when the object will hit the ground. Using the quadratic formula,$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$we find the positive solution that makes physical sense for the problem (time cannot be negative). Plugging in the values,$$t = \frac{-48 \pm \sqrt{(48)^2 - 4(-16)(160)}}{2(-16)}.$$Solving this gives two solutions, but the positive one, which is approximately 5 seconds, is the physically meaningful solution for when the object will hit the ground.

Solve the Problem h=12gt2+vot+ho.h = \frac { 1 } { 2 } g t ^ { 2 } + v _ { o } t + h _ { o }.h=21​gt2+vo​t+ho​.

Solve the Problem $h = \frac { 1 } { 2 } g t ^ { 2 } + v _ { o } t + h _ { o }.$