Question

What a projectile 2 Derive the expression
for the horizontal range of the projectile.
Show that a projectile projected at a
angle of 45° covers meximumo horizontal
range
​

Answer 1

Answer:

Explanation:

Range = u^2 sin2A / g

Now for Rmax  , Sin 2A should be Maximum , therefore max value of Sin 2A is 1 i.e. sin 90  

therefore 90 = 2A

A = 45 degrees

Answer 2

Answer:

Ans is 45 degrees

ok............