How far below an initial straight line path will a projectile fall in one second?

Answer 1

Given no air resistance or other forces acting on the projectile, the projectile falls at a rate of ~9.81 meters per second. Given the position equation is at^2 + vt + x, where a is acceleration, v is velocity, x is the starting position, and t is time. Given an initial velocity and time of zero, the object will have moved ~9.81 meters in the first second.

Answer 2

d = (vit) + (1/2)gt2

d = distance

vi = initial velocity = 0m/s

g = acceleration due to gravity = 9.8m/s2

t = time = 1s

d = (0) + (1/2)(9.8m/s2)(1s2) = 4.9m

The projectile will fall vertically 4.9m in 1s.

Answer 3

Assuming the projectile is launched horizontally, the distance it falls below an initial straight line path in one second can be calculated using the formula for free fall: d = 0.5 * g * t^2, where d is the distance, g is the acceleration due to gravity, and t is the time (1 second in this case). Plugging in the values for g (9.81 m/s^2) and t (1 second) will give you the distance the projectile falls below the straight line path in one second.