Question

Show that for a projectile launched at an angle
45', the maximum height of the projectile is
one quarter of the range ?​

Answer 1

Answer:

The proof is given below

Explanation:

Range = u^2 (sin2a)/g

where a = 45 degree

So Range = u^2/g --- 1

Max height = u^2*sin(a)^2/2g

= u^2/(2*2*g)

Since sin(a)^2 =1/2

So Max height = u^2/4g

From 1 Max Height = Range/4