Question

Example 12 The celocity of a projectile when it is at the greatest height is 1215
times its velocity when it is at half of its greatest height. Determine its angle of
projection.​

Answer 1

Answer:

the number in the question is wrong. i think its √2/5

then answer  is 60

Explanation:

at highest point

v=ucosθ

at half of max height

v(x)=ucosθ

v(y)=usinθ-gt

   v(y)^2-u(y)^2=-2gh/2

v(y)^2=u^2sin^2θ-gu^2sin^2θ/2g=usinθ/√2

now v=√v(x)^2+v(y)^2

v=√u^2cos^θ+u^2sin^θ/2

given that v1/v2=√2/5

tanθ=√3

θ=60