Answer:
y = 0.99 m
Step-by-step explanation:
This is a projectile launching exercise, let's start by finding the components of the initial velocity, using trigonometry
cos θ = v₀ₓ / v₀
sin θ = v_{oy} / v₀
v₀ₓ = vo cos θ
v_{oy} = I go sin θ
v₀ₓ = 15 cos 30 = 12.99 m / s
v_{oy} = 15 sin 30 = 7.5 m / s
Let's find the time it takes to travel x = 18 m
x = v₀ₓ t
t = x / v₀ₓ
t = 18 / 12.99
t = 1,385 s
at this point it is at a height of
y = v_{oy} - ½ g t²
y = 7.5 1.385 - ½ 9.8 1.385²
y = 0.99 m
therefore the camera must place the foot 99 cm from the ground