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a stone is thrown with a velocity of 20m/s at an elevation of angle A, given by tan A = 3/4, what horizontal distance does it cover in 2 sec, and what is its height then above the horizontal plane through the point of projection?

solution:

r(right) Ux=VcosA=(4/5)V

r(down) Uy=VsinA=(3/5)V

R(right)distance=speed x time

height=(3/5)x V x t =(3/5) 20 x t =12 x 2 = 24 m (BUT answer is 4 m??)

distance= (4/5)Vt =(4/5)20t =16t =16x2 =32 m (correct approach??)

sekling
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1 Answers1

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For the height you are expected to include gravity, so there should be a term $-\frac 12gt^2$. For the horizontal distance you are fine.

Ross Millikan
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  • height=(3/5)x V x t −0.5gt^2 =(3/5) 20 x t - 19.4 =4.4 m – sekling Nov 02 '14 at 15:55
  • You have two mistakes in the last. It looks like you are taking $g=9.7 m/s^2$, when $9.8$ is closer and your answer sheet expects $10$. Second, your first term is not calculated correctly if it is $23.8$ as it appears. – Ross Millikan Nov 02 '14 at 15:59
  • with g = 9.8 m/s2 height=(3/5)x V x t −0.5gt^2 =(3/5) 20 x 2 - 0.5x9.8x4 = 24 - 19.6 = 4.4 m

    with g = 10 m/s2 height=(3/5)x V x t −0.5gt^2 =(3/5) 20 x 2 - 0.5x10x4 = 24 - 20 = 4.0 m

     – sekling Nov 02 '14 at 16:17
  • Thanks for your highlights – sekling Nov 02 '14 at 16:20