標題:
projectile motion(3)
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00171827/o/701109110083613873456650.jpg please solve it.
最佳解答:
Let the distance along the slope between the projecting point and the hiting point be L. For the horizontal motion: L.cos(45) = 15t where t is the time of flight i.e. t = L.cos(45)/15 --------------------- (1) For the vertical motion, use equation: s = ut + (1/2)at^2 with s = -L.sin(45), u = 0 m/s, a = -g(=-10 m/s2), t = L.cos(45)/15 hence, -L.sin(45) = (1/2)(-10).[L.cos(45)/15]^2 solve for L gives L = 63.64 m Therefore, the rock lands at a distance of 63.64 m down the slope. Substitute the value of L into equation (1) time of flight t = 63.64.cos(45)/15 s = 3 s
projectile motion(3)
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00171827/o/701109110083613873456650.jpg please solve it.
最佳解答:
Let the distance along the slope between the projecting point and the hiting point be L. For the horizontal motion: L.cos(45) = 15t where t is the time of flight i.e. t = L.cos(45)/15 --------------------- (1) For the vertical motion, use equation: s = ut + (1/2)at^2 with s = -L.sin(45), u = 0 m/s, a = -g(=-10 m/s2), t = L.cos(45)/15 hence, -L.sin(45) = (1/2)(-10).[L.cos(45)/15]^2 solve for L gives L = 63.64 m Therefore, the rock lands at a distance of 63.64 m down the slope. Substitute the value of L into equation (1) time of flight t = 63.64.cos(45)/15 s = 3 s
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