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 32 feet per second,releases a sandbag at the instant it is 12 feet above the ground.At what velocity will it hit the ground? Round your answer to three decimal places.
A)-60.284 ft/sec
B)-42.332 ft/sec
C)-15.071 ft/sec
D)64.985 ft/sec
E)74.332 ft/sec

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 32 feet per second,releases a sandbag at the instant it is 12 feet above the ground.At what velocity will it hit the ground? Round your answer to three decimal places. ​
A)-60.284 ft/sec 
B)-42.332 ft/sec 
C)-15.071 ft/sec 
D)64.985 ft/sec 
E)74.332 ft/sec

Quizplus · Accepted Answer

We can use the formula for the height h(t) of an object thrown upward from a point h_0 with an initial velocity v_0:

h(t) = -16t^2 + v_0t + h_0

For the balloon, we have:

h(t) = 12 + 32t - 16t^2 (note that we use 12 instead of 11, since the sandbag is released when the balloon is 12 feet above the ground)

To find the time at which the sandbag hits the ground, we set h(t) = 0 and solve for t:

0 = 12 + 32t - 16t^2
0 = -2t^2 + 2t + 3
t = (-2 ± sqrt(4 - 4(-2)(3))) / (-4)
t = (1 ± sqrt(7)) / 2

Since we're interested in the negative root (when the sandbag is coming down), we have:

t = (1 - sqrt(7)) / 2

Now we can find the velocity of the sandbag at this time. We can use the derivative of h(t) with respect to t to find the velocity:

v(t) = h'(t) = -32t + v_0

At t = (1 - sqrt(7)) / 2, we have:

v = -32((1 - sqrt(7)) / 2) + 0
v = 16sqrt(7) - 32

Rounding to three decimal places, we get:

v = -42.332 ft/sec

So the answer is B, the balloon will hit the ground with a velocity of -42.332 ft/sec.

The Height Above the Ground of an Object Thrown Upward