A projectile is launched beneath the straight line path it would follow if there was no gravity. How many metres does it fall below the line if it has been travelling for 1 second? 2 seconds?

hfell=1/2 g t^2

To solve this problem, we need to consider that the projectile's motion consists of two independent components: horizontal motion with a constant velocity due to no force acting in that direction, and vertical motion influenced by the force of gravity.

Let's assume that the projectile is launched with an initial velocity (v₀) at an angle above the horizontal path. To determine how many meters the projectile falls below the straight line path at a given time, we first need to find the vertical displacement.

The vertical displacement (Δy) can be found using the equation: Δy = (1/2) * g * t², where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time of flight.

For 1 second of travel:
Δy = (1/2) * 9.8 * (1)²
Δy = 4.9 meters

For 2 seconds of travel:
Δy = (1/2) * 9.8 * (2)²
Δy = 19.6 meters

Therefore, after 1 second of travel, the projectile falls 4.9 meters below the straight line path, and after 2 seconds of travel, it falls 19.6 meters below the line.