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In Fig. , a ball is launched with a velocity of magnitude 10.0 m/s, at an angle of 50.0^@to the horizontal. The launch point is at the base of a ramp of horizontal length d_1 = 6.00 m and height d_2 = 3.60 m. A plateau is located at the top of the ramp. (a) Does the ball land on the ramp or the plateau? When it lands, what are the (b) magnitude and (c) angle of its displacement from the launch point? |
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Answer» Solution :(a) Let m = `d_2//d_1 = 0.600` be the slope of the ramp ,so y = mx there , we choose our coordinate ORIGIN at the point of launch and use eq. thus ` y = tan (50.0^@) x- ((9.80 m//s^2)x^2)/(2(10.0 m//s)^2 (cos 50.0^@)^2) = 0.600 x ` which yield x = 4.99 m. this is less than `d_1` so the ball does land on the ramp (b) Using the value of x FOND in part (a) , we obtain`y = mx = 2.99 m `. thus , the pythagorean theorem yield a displacement MAGNITUDE of `sqrt( x^2 + y^2) = 5.82 m` (c ) The angle is , of course , the angle of the ramp : `tan^(-1) (m)` = 31.0^@`. |
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