InterviewSolution
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InterviewSolution
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A ball is shot from the ground into the air. At the height of 9.1 m, its velocity is v = (7.6 m//s)hati + (6.1 m//s)hatj, withhat ihorizontal and j upward. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball travel? What are the (c) magnitude and (d) angle (below the horizontal) of the ball's velocity just before it hits the ground? |
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Answer» Solution :We DESIGNATE the given velocity `v = (7.6 m//s)hati + (6.1 m//s)hatj` as `vecv_1`, as opposed to the velocity when it reaches the max height, `vecv_2` or the velocity when it returns to the ground `vecv_3`, and take y, as the launch velocity, `vecv_0`as usual. The origin is at its launch point on the ground. Different approaches are available , but since it will be be useful (for the rest of the problem ) to FIRST find the initial y velocity , that is how we will proceed . using eq. we have . `v__(1_y)^2 = (v_0y)^2 -2g DELTA y` substituting values , we have `(6.1 m//s)^2 = v_(0y)^2- 2 ( 9.8 m//s^2) (9.1 m)` Which yields `v_(0y) = 14.7 m//s` . Knowing that `v_(2y)` must equal 0 , we use eq. again but now with `Delta y = h` for the maximum height: `v_(2y)^2 = v_(0y)^2 - 2gh rArr 0 = (14.7 m//s)^2 - 2 (9.8 m//s^2) h` which yield h = 11 m (b) Recalling the derivation of eq. but using `v_(0y)` for `v_0 sin theta_0` and `v_(0x) cos theta_0` we have. `0 = v_(0y) t - 1/2 "gt"^2 , R = v_(0x) t` Which leads to` R = 2v_(0x) v_(0y)lg ` . noting that `v_(0x) = v_(1x) = 7.6 m//s`, we plug in values and obtain `R = 2 (7.6 m//s) (14.7 m//s) (9.8 m//s^2) = 23m` (c ) since `v_(3x) = v_(1x) = 7.6 m//s " and " v_(3Y) = -v_(0y) = -14.7 m//s` , we have `v_3 = sqrt(v_(3x)^2+ v_(3y)^2) = sqrt( (7.6 m//s)^2 + (-14.7 m//s)^2) = 17 m//s` (d) The angle (measured from horizontal for `vecv_3` is one of these possibilities : `tan^(-1) ((-14.7 m)/(7.6m)) = - 63^@ or117^@` |
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