標題:
physics
發問:
A ball,as shown in figure is profected with an initial speed of 30m/s at an angle of 60^o above the horizontal . the ball hits the top of the building of height h, 5s after launching.a. determine the height h of the buildingb.what is the speed of the ball just before hitting on the buildingc.find out... 顯示更多 A ball,as shown in figure is profected with an initial speed of 30m/s at an angle of 60^o above the horizontal . the ball hits the top of the building of height h, 5s after launching. a. determine the height h of the building b.what is the speed of the ball just before hitting on the building c.find out the maximum height H reached above the ground. 更新: in horizontal, v=(30)(cos60)唔係不變GE咩?? 點解B又要搵
最佳解答:
(a) Consider the vertical components, u = 30.sin(60) m/s, t = 5 s, a = -g(=-10 m/s2), s = h using s = ut + (1/2).at^2 h = 30.sin(60) x 5 + (1/2)(-10)(5x5) m = 4.9 m (b) u = 30.sin(60) m/s, t = 5s, a = -10 m/s2, v=? using v = u + at = 30.sin(60) - 10x5 m/s = -24 m/s hence, speed = square-root[24^2 + (30.cos(60))^2] m/s (c) u = 30.sin(60) m/s, a = -10 m/s2, v = 0 m/s, s = H using v^2 = u^2 + 2.a.s 0 = [30.sin(60)]^2 + 2.(-10).H hence, H = [30.sin(60)]^2/20 m
physics
發問:
A ball,as shown in figure is profected with an initial speed of 30m/s at an angle of 60^o above the horizontal . the ball hits the top of the building of height h, 5s after launching.a. determine the height h of the buildingb.what is the speed of the ball just before hitting on the buildingc.find out... 顯示更多 A ball,as shown in figure is profected with an initial speed of 30m/s at an angle of 60^o above the horizontal . the ball hits the top of the building of height h, 5s after launching. a. determine the height h of the building b.what is the speed of the ball just before hitting on the building c.find out the maximum height H reached above the ground. 更新: in horizontal, v=(30)(cos60)唔係不變GE咩?? 點解B又要搵
最佳解答:
(a) Consider the vertical components, u = 30.sin(60) m/s, t = 5 s, a = -g(=-10 m/s2), s = h using s = ut + (1/2).at^2 h = 30.sin(60) x 5 + (1/2)(-10)(5x5) m = 4.9 m (b) u = 30.sin(60) m/s, t = 5s, a = -10 m/s2, v=? using v = u + at = 30.sin(60) - 10x5 m/s = -24 m/s hence, speed = square-root[24^2 + (30.cos(60))^2] m/s (c) u = 30.sin(60) m/s, a = -10 m/s2, v = 0 m/s, s = H using v^2 = u^2 + 2.a.s 0 = [30.sin(60)]^2 + 2.(-10).H hence, H = [30.sin(60)]^2/20 m
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