a) A hot air balloon is traveling vertically up- ward at a constant speed of 4 m

Question

a) A hot air balloon is traveling vertically up- ward at a constant speed of 4 m/s. When it is 27 m above the ground, a package is released from the balloon.
After it is released, for how long is the package in the air? The acceleration of gravity is 9.8 m/s2 . Answer in units of s.

b) What is its speed just before impact with the ground? Answer in units of m/s.

c) Now assume the hot air balloon is traveling vertically downward at a constant speed of 4 m/s. After the package is released, how long is it in the air? Answer in units of s.

d) What is its speed just before impact with the ground? Answer in units of m/s.

Explanation / Answer

v0=4 m/s ; y = -27 m ( below the origin ) ;g =9.8 m/s2

A) y = v0t - 0.5gt2

  -27 = 4t - 0.5.(9.8)t2 solve this equation we get t =2.79 s

B) vt = v0 -gt = 4 -9.8(2.79) = -23.342 m/s (down ward )

C ) v0 =-4 m/s ( down ward )

    y = v0t - 0.5gt2

   -27 = -4t - 0.5(9.8)t2  solve this equation we get t = 1.97 s

D) vt = v0 -gt = -4 - (9.8.1.97) = - 23.306 m/s (down ward )

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