A driver is running 3.6 m/s and dives out horizontally from the edge of a vertical cliff and reaches the water below 2.0s later. A) How high was the cliff? B) How far from its base did the driver hit the water?
h= 1/2 g t^2 figure that.
how far did he go in t seconds? 3.6t
To solve this problem, we need to use the principles of motion and the equations of motion. Let's break it down into two parts:
A) To find the height of the cliff:
We can use the equation of motion: h = ut + 0.5 * g * t^2
where:
h = height of the cliff
u = initial vertical velocity (which is 0 since the driver dives horizontally)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken
Given that t = 2.0s, we can substitute these values into the equation:
h = 0 + 0.5 * (9.8) * (2.0)^2
h = 0 + 0.5 * 9.8 * 4.0
h = 0 + 19.6
h = 19.6 meters
Therefore, the height of the cliff is 19.6 meters.
B) To find the horizontal distance from the base of the cliff to the point where the driver hits the water:
We can use the equation of motion for horizontal distance: d = v * t
where:
d = horizontal distance
v = horizontal velocity (which is 3.6 m/s)
t = time taken which is given as 2.0s
Substituting these values into the equation:
d = 3.6 * 2.0
d = 7.2 meters
Therefore, the driver hits the water 7.2 meters away from the base of the cliff horizontally.