A man is in a hot air balloon that takes off from rest and rises with a constant

Question

A man is in a hot air balloon that takes off from rest and rises with a constant 0.35 m/s^2 acceleration. Shortly after take off the man notices that he forgot his camera. A "friend" throws the camera up to him. If the friend throws the camera 12 s after takeoff, with what minimum speed must the friend throw the camera in order for the man to be able to catch it? Ignore air resistance and assume that the camera is thrown, essentially, from ground level.

Explanation / Answer

so inital a balloon=.35m/s2 V of balloon=0m/sec at t=12 h=0+.5at2 h=.5*.35*12^2 h=25.2m v of balloon at t=12 V=at=.35*12=4.2m/sec so distance between camera and balloon is 25.2 m let the velocity of camera be v m/sec and a=-9.8m/s2 Vrelative of cam wrt to balloon is V-4.2 a relative of cam wrt balloon -9.8-.35=-10.15m/s2 h=25.2 h=ut+.5at^2 25.2=vt-(4.2)t+.5*-10.15*t^2 v=25.2/t +4.2+5.075t for v to be min dv/dt=-25.2/t^2+5.075=0 =>t=sqrt(25.2/5.075)=2.2283sec d2v/dt2=50.4/t^3 d2v/dt2(t=2.2283)=4.55>0 at t=2.2283 v is minimum Vmin=26.82m/sec

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