1 Answer | Add Yours
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.
We’ve answered 317,821 questions. We can answer yours, too.Ask a question