Kinematics Question 636
Question: An object it’s projected with a velocity of $ 20m/s $ making an angle of $ 45{}^\circ $ with horizontal. The equation for the trajectory it’s $ h=Ax-Bx^{2} $ where h it’s height, x it’s horizontal distance, A and B are constants. The ratio A: b is $ ( \text{g = 10 m}{{\text{s}}^{-2}} ) $
Options:
A) $ 1:5 $
B)$ 5:1~~ $
C) $ 1:40 $
D)$ 40:1 $
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Answer:
Correct Answer: D
Solution:
[d] Given $ h=Ax-\text{B}{{\text{x}}^{\text{2}}} $ ,
on comparing with $ y=\text{x}\tan \theta -\frac{gx^{2}}{2u^{2}{{\cos }^{2}}\theta } $ ,
we get $ A=\tan \theta =tan45{}^\circ =1, $
and $ \text{B}\text{=}\frac{g}{2u^{2}{{\cos }^{2}}\theta }=\frac{10}{2\times 20^{2}\times {{\cos }^{2}}45{}^\circ }=\frac{1}{40} $
$ \therefore \frac{\text{A}}{\text{B}}=40. $
BETA
