Question

Find the angle of projection for a projcetile motin whose range is n times to the h=maximum height

Answer 1

the correct answer is (theta) = tan^(-1) 4/n
How? (Solution below)

Let (theta) be the angle of projection.
R (range) = [2(u^2) sin(theta) cos(theta)]/(g) ...(eq1)
H (max ht.) = [(u^2) sin^2(theta)]/(2g) ...(eq2)

Also, R = n (H)
Now, by putting the values from eq 1 n 2 in this we get,

[2(u^2) sin(theta) cos(theta)]/(g) = n [(u^2) sin^2(theta)]/(2g)

[2 sin(theta) cos(theta)] = n [sin^2(theta)]/(2)

[4 sin(theta) cos(theta)] = n [sin(theta) sin(theta)]

[4 cos(theta)] = n [sin(theta)]

4 = n [tan(theta)]

tan(theta) = 4/n

(theta) = tan^(-1) 4/n