A rock is dropped from a riverbank that is 4 feet above the surface of the river. After how many seconds does the rock hit the surface of the water? At highest?
for non-metric units, the height h is given by
h = -16t^2 + 4, where h is in feet above the river level, and t is in seconds
so we want h = 0
0 = -16t^2 + 4
16t^2 = 4
t^2 = 1/4
t = 1/2
it takes 1/2 second
To find the time it takes for the rock to hit the surface of the water, we can use a formula called "the equation of motion" which relates distance, time, and acceleration due to gravity. The equation is:
d = 1/2 * g * t^2
where:
d is the distance (in this case, the height of the riverbank) = 4 feet
g is the acceleration due to gravity = 32 feet/s^2 (approximately)
t is the time (in seconds) we want to find
In this case, since we are dropping the rock vertically downward, the initial velocity is 0. This means we can simplify the equation to:
d = 1/2 * g * t^2
Now, plug in the known values:
4 = 1/2 * 32 * t^2
To solve for t^2, divide both sides by 16:
4/16 = (1/2 * 32 * t^2)/16
1/4 = t^2/2
Multiply both sides by 2:
2 * 1/4 = 2 * (t^2/2)
1/2 = t^2
Take the square root of both sides:
√(1/2) = √(t^2)
t = √(1/2)
So, the rock will hit the surface of the water after approximately √(1/2) seconds.