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phy-motion(4)
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Agolf ball is struck from a point O with velocity 24 ms^-1 at an angle of 40° to the horizontal. The ball firsthits the ground at a point P, which is at a height h metres above the level ofO.圖片參考：http://img191.imageshack.us/img191/1149/13186336.jpgThehorizontal distance between O and P is 57 metres.a) Show that... 顯示更多 Agolf ball is struck from a point O with velocity 24 ms^-1 at an angle of 40° to the horizontal. The ball firsthits the ground at a point P, which is at a height h metres above the level ofO. 圖片參考：http://img191.imageshack.us/img191/1149/13186336.jpg Thehorizontal distance between O and P is 57 metres.a) Show that the time that the balltakes to travel from O to P is 3.10 s, correct to 3 sigfig.b) Find the value of hc) Find the speed with which theball hits the ground at Pd) Find the angle between thedirection of motion and the horizontal as the ball hits the ground at P
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phy-motion(4)
發問:
Agolf ball is struck from a point O with velocity 24 ms^-1 at an angle of 40° to the horizontal. The ball firsthits the ground at a point P, which is at a height h metres above the level ofO.圖片參考：http://img191.imageshack.us/img191/1149/13186336.jpgThehorizontal distance between O and P is 57 metres.a) Show that... 顯示更多 Agolf ball is struck from a point O with velocity 24 ms^-1 at an angle of 40° to the horizontal. The ball firsthits the ground at a point P, which is at a height h metres above the level ofO. 圖片參考：http://img191.imageshack.us/img191/1149/13186336.jpg Thehorizontal distance between O and P is 57 metres.a) Show that the time that the balltakes to travel from O to P is 3.10 s, correct to 3 sigfig.b) Find the value of hc) Find the speed with which theball hits the ground at Pd) Find the angle between thedirection of motion and the horizontal as the ball hits the ground at P
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(a) Horizontal velocity component at O = 24cos(40) ms/ = 18.385 m/sHence, time taken = 57/18.385 s = 3.10 s (b) Consider the vertical motion of the ball. Use equation of motion: s = ut + (1/2)at^2 with u = 24.sin(40) m/s = 15.43 m/s, t = 3.1s, a = -g(=-9.8 m/s2), s = hhence, h = (15.43 x 3.1 + (1/2).(9.8).(3.1)^2) m = 0.744 m (c) Kinetic energy at O = (m/2).(24)^2 J where m is the mass of the golfballEnenrgy at P = (m/2)v^2 + mg(0.744) J where v is the speed at PBy conservation of energy, (m/2).(24)^2 = (m/2)v^2 + mg(0.744) i.e. v = 23.69 m/s (d) Let a be the angle,cos(a) = 18.385/23.69 a = 39 degrees其他解答:
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