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katykitz
pls help... for a projectile launched from ground at an angle theta with the vertical , the maximum height reached is a. directly proportional to sin^2θ b. directly proportional to cos^2θ
b. If the projectile is launched with an angle theta with the horizontal then only the answer is a. The formula to calculate height is: h= {(u^2)(sin^2θ)}/2g ; where u= initial velocity, g=acceleration due to gravity, θ = angle of projectile with the horizontal.
a. directly proportional to sin^2θ
Lets see - if it was proportional to cos^2(theta) then a projectile launched vertically should have reached the least distance - which clearly is not true. Hence, sin^2(theta). It helps sometimes to put in 0 and 90 as the values of angles.
b, becs, the vertical component of the velocity would be V cos(θ), and by applying the eq. of motion, v^2-u^2=2(a)(h) where, v=0, u= V cos(θ) &a=(-g), we find that the height, h is directly proportional to cos^2(theta). @ankurchiitd: Cos(0)=1 is the max. value for the cosine function, and when the projectile would be launched vertically, i.e. with 0 degree to the vertical, it would attain the max. height.
Note that incorrect answers are not considered abuse. If you see an incorrect answer, do as you did and simply correct it.
umm...guys tnx 4 ur help...but could some1 pls tell me the exact answer...its so confusing !!!
Several people posted the exact answer :)
which is it?? ....a? pls tell me!
Check out the answer that has two medals... Vs the ones that have none ;)
k ...got it...tnx a lot.. :)