Solve the problem.
-Consider the quadratic model $h ( t ) = - 16 t ^ { 2 } + 40 t + 50$ for the height (in feet), $h$, of an object $t$ seconds after the object has been projected straight up into the air. Find the maximum height attained by the object. How much time does it take to fall back to the ground? Assume that it takes the same time for going up and coming down.
A)maximum height = 75 ft; time to reach ground = 2.5 seconds
B)maximum height = 50 ft; time to reach ground = 1.25 seconds
C)maximum height = 75 ft; time to reach ground = 1.25 seconds
D)maximum height = 50 ft; time to reach ground = 2.5 seconds

Question

Quizplus · Accepted Answer

The maximum height can be found using the vertex formula for a quadratic equation, $h(t) = -16t^2 + 40t + 50$. The time at maximum height, $t_{max}$, is given by $-b/(2a) = -40/(2 \times -16) = 1.25$ seconds. Plugging $t_{max}$ back into the equation gives the maximum height, $h(1.25) = -16(1.25)^2 + 40(1.25) + 50 = 75$ feet. The object takes 1.25 seconds to reach the maximum height and another 1.25 seconds to fall back down, totaling 2.5 seconds to reach the ground.

Solve the Problem h(t)=−16t2+40t+50h ( t ) = - 16 t ^ { 2 } + 40 t + 50h(t)=−16t2+40t+50

Solve the Problem $h ( t ) = - 16 t ^ { 2 } + 40 t + 50$