In the figure, a ball is launched with a velocity of magnitude 6.00 m/s, at an a

Question

In the figure, a ball is launched with a velocity of magnitude 6.00 m/s, at an angle of 49.0? to the horizontal. The launch point is at the base of a ramp of horizontal length d1 = 6.00 m and height d2 = 3.60 m. A plateau is located at the top of the ramp. (a) Does the ball land on 1 - the ramp or 2 - the plateau? (Give the number of the correct answer.) When it lands, what are the (b) magnitude and (c) angle of its displacement from the launch point? Can you please explain how you got to your conclusion so i can get a better understanding

Explanation / Answer

Given v=9.00m/s, ?=50°, d1=4.10m, d2=3.60m, g=9.81m/s² ramp angle a = atan(d2/d1) = 41.3° ramp length L = v(d1²+d2²) = 5.46m The max height of a projectile is given by: h = (vsin?)²/2g = 2.42m [so ball won't clear ramp] The range on even ground (w/o ramp) would be: R = v²sin(2?)/g = 8.13m So the range on the ramp will be: Rs = R(1-tana/tan?)/cosa = 2.84m

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