A package is dropped from a helicopter moving upwards at 15 m/s. If it takes 16.

Question

A package is dropped from a helicopter moving upwards at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released? A car is travelling north velocity at 17.7 m/s. After 12 s. its velocity is 14.1 m/s in the same direction. Find the magnitude and direction of the car's average acceleration. Vector A has length 4 units and directed to the north. Vector B has length 9 units and is directed to the south. Calculate the magnitude and direction of A + B Vector A points to the north and has length A. Vector B points to the east and has length B = 2.0 A. Find the magnitude of C = 4A = B in terms of A.

Explanation / Answer

5)

T = sqrt(2*h/g)

h = T2*g/2 = 16*16*9.8/2 = 1254.4 m

6)

total displacement = 6.7 - 1.2 = 5.5 miles east

avg velocity = 5.5/6.1 = 0.902 miles/hour

7)

v = u + at

14.1 = 17.7 + a*12

=> a = - 3.6/12 = - 0.3 m/s2

direction of acceleration is south

8)

A = 4j, B =-9j

A + B = -5j

magnitude = 5

direction is south

9)

A = Aj, B = 2Ai

C = 4Aj + 2Ai

magnitude of C = A*sqrt(16+4) = 4.472*A

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