projectiles: how do I solve a range question using kinematics for the x and y components instead of the range equation?

Answer 1

Kavitha

If initial velocity is v, angle of elevation θ, initial horizontal velocity is v cos θ, vertical velocity v sin θ.

s = ut + 0.5 a t ²

Lets assume positive direction is upwards

So projectile is on the ground when s = 0

u = v sin θ

a = g (acceleration due to gravity) = 9.8 m/s/s

So projectile on ground when s=0, ie when vt sin θ - 4.9 t² = 0

So t=0 or v sin θ = 4.9 t, so t = (v sin θ / 4.9)

So the projectile hits the ground after this time.

The horizontal speed is constant at v cos θ, because no net force in horizontal direction.

So range will be v cos θ x (v sin θ / 4.9) = v² sin θ cos θ / 4.9

Mike