A burglar drops a bag of loot from a window in a hotel. The bag takes 0.15 secon

Question

A burglar drops a bag of loot from a window in a hotel. The bag takes 0.15
seconds to pass the 1.6 m tall window of your room as it falls toward the ground.
How far above the top of your window is the burglar who dropped the bag? (The
bag’s initial speed is zero.) The same burglar goes to another hotel and throws a
sack of loot vertically downward with an initial speed of 12 m/s from the roof, 30.0
m above the ground. (a) How long does it take the sack to reach the ground? (b)
What is the speed of the sack at impact?

Explanation / Answer

Time t = 0.15 s

Let v be the speed of the bag at the top of the window,From kinematic relation

                            1.6 = u (0.15) + [0.5*9.8*(0.15)^2]

                            1.6 = 0.15 u + 0.11

                               u = 9.93 m/s

   

Let H be the Height above the top of your window is the burglar who dropped the bag

H = (9.93)^2/(2*9.8) = 5.034 m  

(a)

S = 30 m

u = 12 m/s

From kinematic relation

       S = ut + (1/2)gt^2

30 = 12 t + 0.5*9.8*t^2

4.9 t^2 + 12 t - 30 = 0

On solving the above quadratic equation we get

    time taken t = 1.536 s

(b)

Velocity v = 12 + (9.8*1.536)

                = 12 + 15.05

                 = 27.05 m/s

                                                                                              

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