1. A rock is projected from the edge of the top of a building with an initial ve
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Question
1. A rock is projected from the edge of the top of a building with an initial velocity of 40 ft/s at an angle of 53 degrees above the horizontal. The rock strikes the ground a horizontal distance of 82 ft from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?(the answer is 77ft but I don't know how to solve it.)
2. A baseball leaves a bat at an angle of 30 degrees above the horizontal. The ball strikes a fence that is 100m horizontally from the batter at a height of 5.0 m above the height of the bat when it struck the ball. What was the speed of the ball as it left the bat. (the answer is 35 m/s but I don't know how to solve it)
3. A ferry boat is sailing at 12mph 30 degrees W of N with respect to a river that is flowing at 6.0 mph E. As observed from the shore, the ferry boat is sailing: ( the answer is 30 degrees W of N but I don't know how to solve it).
Thanks so much in advance
Explanation / Answer
2. Given that the angle of projection is = 30o Horizontal displacement is X = 100 m Vertical displacement is Y = 5.0 m -------------------------------------------------- Initial horizontal velocity is Ux = U cos Initial vertical velocity is Uy = U sin Horizontal displacement is X = Ux*t t = X / Ux Vertical displacement is Y = Uy*t - (1/2)gt2 Y = Uy*(X / Ux) - (1/2)g(X/ Ux)2 Y = U sin*(X / Ucos) -(1/2)g( X / U cos)2 Y = tan*X - g X2 / 2(U cos)2 g X2 / 2(U cos)2 = X tan -Y U2 = g X2 / 2(cos)2 (X tan - Y) = (9.8 m/s2)(100 m)2 / 2(cos 30o) (100 tan30o - 5.0 m ) = 1072.91 Then we get U = 32.75 m/s Horizontal displacement is X = 100 m Vertical displacement is Y = 5.0 m -------------------------------------------------- Initial horizontal velocity is Ux = U cos Initial vertical velocity is Uy = U sin Horizontal displacement is X = Ux*t t = X / Ux Vertical displacement is Y = Uy*t - (1/2)gt2 Y = Uy*(X / Ux) - (1/2)g(X/ Ux)2 Y = U sin*(X / Ucos) -(1/2)g( X / U cos)2 Y = tan*X - g X2 / 2(U cos)2 g X2 / 2(U cos)2 = X tan -Y U2 = g X2 / 2(cos)2 (X tan - Y) = (9.8 m/s2)(100 m)2 / 2(cos 30o) (100 tan30o - 5.0 m ) = 1072.91 Then we get U = 32.75 m/s Y = Uy*(X / Ux) - (1/2)g(X/ Ux)2 Y = U sin*(X / Ucos) -(1/2)g( X / U cos)2 Y = tan*X - g X2 / 2(U cos)2 g X2 / 2(U cos)2 = X tan -Y U2 = g X2 / 2(cos)2 (X tan - Y) = (9.8 m/s2)(100 m)2 / 2(cos 30o) (100 tan30o - 5.0 m ) = 1072.91 Then we get U = 32.75 m/s Y = tan*X - g X2 / 2(U cos)2 g X2 / 2(U cos)2 = X tan -Y U2 = g X2 / 2(cos)2 (X tan - Y) = (9.8 m/s2)(100 m)2 / 2(cos 30o) (100 tan30o - 5.0 m ) = 1072.91 Then we get U = 32.75 m/sRelated Questions
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