Question 286912: Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)
A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 48 feet above the ground.
(a) How many seconds after its release will the bag strike the ground? (Round your answer to two decimal places.)
(b) At what velocity will it strike the ground? (Round your answer to three decimal places.)
Answer by dabanfield(803)   (Show Source):
You can put this solution on YOUR website! Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)
A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 48 feet above the ground.
(a) How many seconds after its release will the bag strike the ground? (Round your answer to two decimal places.)
(b) At what velocity will it strike the ground? (Round your answer to three decimal places.)
a.) We can use the equation d = v(0) + (1/2)a*t^2 where v(0) is the initial velocity. In this case d = 48, v(0) = 0 and a = 32. Substitute these values and solve for t.
b.) We can use the equation v(f) = v(0)+ a*t where v(f) is the final velocity and v(0) is the initial velocity. In this case v(0) = 0, a = 32. Substituting these values and the one calculated for t in part a.) you can calculate v(f) which is the velocity when the sandbag hits the ground.
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