A rock thrown horizontally from a bridge hits the water below 30 meters away in the horizontal direction. If
the rock was in the air for 2 seconds, how tall is the bridge?
Use the following equation; the horizontal distance is irrelevant.
d=Vi+1/2at^2
Where
vi=0m/s ** there is no initial velocity in the y-direction.
a=g=9.8m/s^2
and
t=2s
Solve for d:
d=0+1/2(9.8m/s^2)(2s)^2
d=19.6.m=20m
To determine the height of the bridge, we can use the equations of motion in projectile motion. We know that the rock was thrown horizontally, which means its initial vertical velocity (Vy) is 0 m/s.
The time of flight (t) is given as 2 seconds, and we are given the horizontal distance traveled (d) as 30 meters.
Using the equation for horizontal distance (d = Vx * t), where Vx is the initial horizontal velocity, we can find Vx:
30 meters = Vx * 2 seconds
Vx = 30 meters / 2 seconds
Vx = 15 meters per second
Since Vy is 0 m/s, the initial velocity (V) of the rock can be found using the Pythagorean theorem:
V^2 = Vx^2 + Vy^2
V^2 = 15 meters per second^2 + 0 meters per second^2
V^2 = 225 meters^2 per second^2
V = √225 meters per second
V = 15 meters per second
To find the height (h) of the bridge, we can use the equation for vertical distance (h = Vy * t + 0.5 * g * t^2), where g is the acceleration due to gravity (approximately 9.8 m/s^2):
h = 0 meters per second * 2 seconds + 0.5 * 9.8 meters per second^2 * (2 seconds)^2
h = 0 meters + 0.5 * 9.8 meters per second^2 * 4 seconds^2
h = 0 + 0.5 * 9.8 meters per second^2 * 16 seconds^2
h = 0 + 0.5 * 9.8 meters per second^2 * 16
h = 0 + 0.5 * 156.8 meters
h = 0 + 78.4 meters
h = 78.4 meters
Therefore, the height of the bridge is approximately 78.4 meters.