Question

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

Answer 1

a

Answer 2

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.

Answer 3

a. directly proportional to sin^2θ

Answer 4

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.

Answer 5

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.

Answer 6

Note that incorrect answers are not considered abuse. If you see an incorrect answer, do as you did and simply correct it.

Answer 7

umm...guys tnx 4 ur help...but could some1 pls tell me the exact answer...its so confusing !!!

Answer 8

Several people posted the exact answer :)

Answer 9

which is it?? ....a? pls tell me!

Answer 10

Check out the answer that has two medals... Vs the ones that have none ;)

Answer 11

k ...got it...tnx a lot.. :)