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The height and distance along horizontal plane of projectile are x=6t,y=8t-5t*t.What is the velocity when velocity is projected?

Asked by sagarikadevnath devnath 1st October 2013, 7:03 PM
Answered by Expert
Answer:

Given,

x = 6t and y = 8t-5t2

Comparing the equation,

  x = 6t with x = (u cos?)*t

? u cos?= 6   (1)

 And,

 y = 8t-5t2 with y =(u sin? )*t (1/2) gt2

 ? u sin? = 8 ..(2)

 Squaring (1) and (2) and adding,

u2cos2?+ u2sin2? = 62 + 82

u2(cos2?+ sin2?)= 36+64

u2 * 1 = 100             [since, cos2?+ sin2? = 1]

u2 = 100

 u = 10 ms1 or 10 m/s

Answered by Expert 1st October 2013, 9:44 PM
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