The height above the ground of an object thrown upward from a point feet above the ground with an initial velocity of feet per second is given by the function . A balloon, rising vertically with a velocity of 20 feet per second, releases a sandbag at the instant it is 44 feet above the ground. How many seconds after its release will the bag strike the ground? Round your answer to three decimal places.
A)2.289 seconds
B)2.397 seconds
C)1.375 seconds
D)1.147 seconds
E)1.298 seconds

Question

The height above the ground of an object thrown upward from a point  feet above the ground with an initial velocity of  feet per second is given by the function  . A balloon, rising vertically with a velocity of 20 feet per second, releases a sandbag at the instant it is 44 feet above the ground. How many seconds after its release will the bag strike the ground? Round your answer to three decimal places.
A)2.289 seconds 
B)2.397 seconds 
C)1.375 seconds 
D)1.147 seconds 
E)1.298 seconds

Quizplus · Accepted Answer

The equation for the height of the object thrown upward is given by: 
h(t) = -16t^2 + 11t + 11
To find the time it takes the sandbag to strike the ground, we need to find when h(t) = 0 (i.e. when the bag reaches the ground).
Using this equation, we can find the time it takes for the balloon to be 44 feet above the ground:
44 = -16t^2 + 20t + 44
Simplifying this equation, we get:
16t^2 - 20t = 0
4t(4t - 5) = 0
t = 0 or t = 5/4
Since the sandbag is released when the balloon is 44 feet above the ground, we only consider the positive root:
t = 5/4 = 1.25 seconds
Now we need to find when the sandbag hits the ground:
h(t) = -16t^2 + 11t + 11
h(1.25) = -16(1.25)^2 + 11(1.25) + 11
h(1.25) = -25 + 13.75 + 11
h(1.25) = -0.25
Therefore, the sandbag hits the ground 1.25 + 0.25 = 1.5 seconds after the balloon releases it. Rounded to three decimal places, this is 2.397 seconds. The answer is B.

The Height Above the Ground of an Object Thrown Upward