Question

pls help ...
For very small angles of projection 'α' with horizontal , the ratio of the range to the maximum height of a particle projected upwards from level ground with initial speed 'u' is
a. 4/α
b. 4u/gα

Answer 1

For small angles, sinα = α
using the range equation
\[range = u ^{2}\sin(2\alpha)/g = u ^{2}2\alpha/g\]
The velocity at max height is 0, so\[height = (u _{f}^{2} - u _{i}^{2})/(-2g) = (usin( \alpha))^{2}/2g = u ^{2} \alpha ^{2}/2g\]
so the ratio is\[(u ^{2}2\alpha/g)/(u ^{2} \alpha ^{2}/2g) = 4/\alpha\]

Answer 2

tnx a lot...i'm sorry i couldnt thank u earlier...i thought this question was not answered..