Question

a projectile is projected with velocity v making an angle theta with the horizontal direction, then the maximum horizontal range _______.

Answer 1

Is 0

(˘･_･˘)

Hopefully uh Understand

A projectile is projected with velocity v making an angle theta with the horizontal direction, then the maximum horizontal range 0

Answer 2

Answer:

$range \: \: max \: = \frac{v^{2} }{g}$

Explanation:

Considering horizontal motion,

Vx = Vcos theta

ax = 0

Sx = R

time = t

R = (v^2 sin2 thata) ÷ g

Sin 2theta = 1

Theta = 45°

[For the max range]

R = v^2/g