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F.5 physics projected motion
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A shell is fired from a point on a horizontal ground 100m from the foot of a verticaltower of height 90m. The angle of projection is adjusted to be 60° Such that the shell can just clear the tower. When the shell reaches its maximum height, it explores into 3 equal fragments. Immediately after explosion,... 顯示更多 A shell is fired from a point on a horizontal ground 100m from the foot of a vertical tower of height 90m. The angle of projection is adjusted to be 60° Such that the shell can just clear the tower. When the shell reaches its maximum height, it explores into 3 equal fragments. Immediately after explosion, one fragment is found to be stationary, one fragment moves upward with a speed of 10m/s. Use g=10m/s^2 a) find the speed of projection of the shell b) find the maximum height reached by the shell Picture: http://postimg.org/image/4x5joambn/ Thank you for help!
最佳解答:
(a) Let U be the speed of projection of the shell. Horizontally, [U.cos(60)].t = 100 where t is the time of flight of the shell from projection point to the top of the tower. Hence, t = 100/[U.cos(60)] Vertically, use equation: s= ut + (1/2)at^2 with s = 90 m, u = U.sin(60), a = -g(=-10 m/s^2), t = 100/[U.cos(60)] hence, 90 = [U.sin(60)].[100/(U.cos(60)] + (1/2).(-10).[100/(U.cos(60)]^2 i.e 90 = 100.tan(60) - 5[10000/U^2.cos^2(60)] solve for U gives U = 49.03 m/s (b) Use equation: v^2 = u^2 + 2a.s with v = 0 m/s, u = 49.03.sin(60) m/s = 42.46 m/s, a = -g(=-10 m/s^2), s =? hence, 0 = 42.46^2 + 2(-10).s s = 90.14 m
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