SOLUTION: A projectile is fired at an angle of 30° to the horizontal with a velocity of 40ms-1. Calculate the velocity attained after 1sec.(g=10ms-2?

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Question 1199411: A projectile is fired at an angle of 30° to the horizontal with a velocity of 40ms-1. Calculate the velocity attained after 1sec.(g=10ms-2?
Answer by ikleyn(52785) About Me  (Show Source):
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A projectile is fired at an angle of 30° to the horizontal with a velocity of 40 m/s.
Calculate the velocity attained after 1 sec.(g = 10 m/s2)
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At the starting moment, vertical velocity component is  V%5Bvert%5D%280%29%5D = 40*sin(30°) = 40%2A%281%2F2%29 = 20 m/s.


The time to get the maximum height is  t%5Bh_max%5D = V%5Bvert%5D%280%29%2Fg = 20%2F10 = 2 seconds,
under the given condition, so after 1 second the projectile is still moving up.



Vertical component of the velocity at t = 1 second is  

    V%5Bvert%5D%281%29 = V%5Bvert%5D%280%29+-+g%2At = 20 - 10*1 = 10 m/s.



Horizontal component of the projectile is  V%5Bhor%5D = 40*cos(30°) = 40%2A%28sqrt%283%29%2F2%29 m/s
remains constant during the entire flight (till hitting the ground). 



Thus the velocity attained after 1 second is


      V(t=1) = sqrt%28%28V%5Bvert%5D%281%29%29%5E2+%2B+%28V%5Bhor%5D%29%5E2%29 = sqrt%2810%5E2+%2B+%2840%2A%28sqrt%283%29%2F2%29%29%5E2%29 = sqrt%28100+%2B+1200%29 = sqrt%281300%29 = 36.055 m/s.   ANSWER

Solved.