I need to make a plan for a stunt man that is hanging off of a cliff at 5 m high. He must land into a car that will be passing underneath (using a constant acceleration). I am stuck on which equation to use that would represent the man dropping into the car


You can use the $s=u_yt+\frac{1}{2} g t^2$ kinematics formula. Let's assume your stuntman just drops from that height . Thus making your time of descent $t=\sqrt{\frac{10}{g}} sec$

Now plan your stunt accordingly .

The horizontal displacement of the car must be planned accordingly so that it reaches the location of drop exactly at $t$ sec (Preferably , a bit before that)

  • $\begingroup$ How will the cars constant acceleration play into this? $\endgroup$ – Jennifer Kuhn Pometo May 14 '17 at 2:14
  • $\begingroup$ $x=\frac{1}{2}a t^2$ assuming the car was initially at rest and $a$ is acceleration of car @JenniferKuhnPometo $\endgroup$ – The Dead Legend May 14 '17 at 2:15
  • $\begingroup$ @JenniferKuhnPometo You may acept the answer if it was helpful to you $\endgroup$ – The Dead Legend May 14 '17 at 2:16

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