標題:
AL_Physics Projectile motion question
發問:
A boby is projected at an angle of 40 degree to the horizontal so as to just clear two walls of equal height 10m , distance 20m apart, Find the range of the projectile, Answer:55m I still can't reach the answer successfully. Noted that it's only a MC-question.
Let u be the initial speed t be the travelling time Consider horizontal component, Horizontal velocity, uH = ucos40? ms-1 So, range of projectile, sH = utcos40? m Consider the range when the projectile rises to the edge of the wall Initial vertical velocity, uV = usin40? ms-1 By equation of motion, vv2 = uv2 – 2gs vv = √(u2sin240? - 200) ms-1 Consider the range that crossing the two walls. Let T be the time required to cross the two walls So, uTcos40? = 20 T = 20 / ucos40? s Once again, by equation of motion, s’ = vvT – 1/2 gT2 0 = 20 / ucos40? √(u2sin240? - 200) – 5(20 / ucos40?)2 Solving, u = 23.6214 ms-1 Now, consider the whole range of the projectile. Initial vertical velocity, uw = 23.6214sin40? = 15.1836 ms-1 By equation of motion, s = uwt – 1/2 gt2 0 = 15.1836t – 5t2 t = 3.0367 s So, range of projectile = 15.1836 X (3.0367)cos40? = 55.0 m
其他解答:
AL_Physics Projectile motion question
發問:
A boby is projected at an angle of 40 degree to the horizontal so as to just clear two walls of equal height 10m , distance 20m apart, Find the range of the projectile, Answer:55m I still can't reach the answer successfully. Noted that it's only a MC-question.
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最佳解答:Let u be the initial speed t be the travelling time Consider horizontal component, Horizontal velocity, uH = ucos40? ms-1 So, range of projectile, sH = utcos40? m Consider the range when the projectile rises to the edge of the wall Initial vertical velocity, uV = usin40? ms-1 By equation of motion, vv2 = uv2 – 2gs vv = √(u2sin240? - 200) ms-1 Consider the range that crossing the two walls. Let T be the time required to cross the two walls So, uTcos40? = 20 T = 20 / ucos40? s Once again, by equation of motion, s’ = vvT – 1/2 gT2 0 = 20 / ucos40? √(u2sin240? - 200) – 5(20 / ucos40?)2 Solving, u = 23.6214 ms-1 Now, consider the whole range of the projectile. Initial vertical velocity, uw = 23.6214sin40? = 15.1836 ms-1 By equation of motion, s = uwt – 1/2 gt2 0 = 15.1836t – 5t2 t = 3.0367 s So, range of projectile = 15.1836 X (3.0367)cos40? = 55.0 m
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