10. A sky diver drops his smuggled 1kg weight from physics lab out of the air pl

Question

10. A sky diver drops his smuggled 1kg weight from physics lab out of the air plane before he jumps, which is flying horizontally going 175 m/s at a height of 2 km. IF WE ASSUME NO AIR RESISTANCE, what is the Velocity (magnitude and direction) when it hits the ground?

11. If an object initially at rest is given a velocity of: vx =-1.93 m/s and vy = 2.51 m/s. What is the magnitude and direction its displacement after a time of 3.50 seconds?

12. A cannon shell is fired on a flat surface and it is found that the shell is in the air for 2.78 s and reaches a distance of 94.7 m. What was the initial Launch angle and speed?

13. In an airport moving walkways help get passengers to their gates faster. In one case a person stands on the walkway and arrives at the gate in 525 seconds. Another time the walkway is broken and he makes the same trip in 357 seconds walking at a constant speed. How fast could he make it, if he walked the same speed on the moving walkway?

Explanation / Answer

Given

10.   mass m = 1kg, position of mass h = 2km = 2000 m from the surface

   due to its position it would have gravitaional potential energy is U = mgh = 1*9.8*2000 = 19600 J

when it was dropped this energy will be converted in to k.e , at the time when it hits the ground tha is

   mgh = 0.5 mv^2

   v = sqrt(2gh)
   = sqrt(2*9.8*2000)
  
   = sqrt(39200) m/s

   = 14.07 m/s

hits the ground normally as there is no air resistance

11.Given

   Vx = -1.93 m/s, Vy = 2.51 m/s

and time t = 3.5
s

   the magnitude of velocity is v = sqrt(vx^2+vy^2)

                   = sqrt((-1.93)^2+(2.51)^2) m/s
                   = 3.166 m/s

the direction is tan(180-theta ) = 2.51/1.93 ==> -tan theta = (2.51/1.93) = 142 degrees

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