A man leaps off a cliff at 12 m/s and an angle 32 degrees to the horizontal. If he has to reach a cliff 6 m away how far below the ledge must his launch point be?



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To approach this problem you must apply the two equations of projectile motion.  Remember we can separate the motion into two separate problems with a horizontal motion solution, and a vertical solution.

The two equations are

Rx = VxT  where Rx is the range (6.0 meters in this case), Vx is the horizontal componant of the velocity [Vcos(theta) = 12 m/s cos(32) = 10.2 m/s in this case], ant T is the time of flight.

H = VyT + 0.5 gT^2  where Vy is the vertical component of the velocity [Vsin(theta) = 12 m/s sin(32) = 6.4 m/s in this case]

To get to the answer we have to solve the first equation for T and the substitute that into the second equation.

T = Rx/Vx = 6.0m/10.2m/s = 0.588 s

H = 6.4m/s(0.588s) + 0.5(-9.80m/s^2)(0.588s)^2 = 2.04 m.

Therefore the launch site must be 2.0 meters lower than the ledge in order to reach it in this manner.


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