A horse runs off of a 11 meter high cliff into the sea. The horse is running at a speed of 3.6 m/s. How much time does the horse spend in the air? How far from the base of the cliff is the horse when it hits the water?
Time in the air is given by:
(1/2) g T^2 = 11 m
V * T = horizontal distance traveled
First, how long in the air?
h=1/2 g t^2 solve for t.
How far away horizontally?
d=rt
d=3.6m/s * time
To calculate the time the horse spends in the air, we need to use the kinematic equation that relates distance, initial velocity, time, and acceleration:
y = y0 + v0t - (1/2)gt^2
In this equation, y is the vertical displacement (the height of the cliff), y0 is the initial vertical position (0 in this case, since the horse is on the cliff), v0 is the initial velocity (3.6 m/s in this case), t is the time spent in the air (what we're trying to find), and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the known values into the equation, we have:
11 = 0 + (3.6)t - (1/2)(9.8)t^2
Simplifying the equation, we get:
(1/2)(9.8)t^2 - (3.6)t + 11 = 0
Now, we can solve this quadratic equation for t. We can either use the quadratic formula or factor the equation. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = (1/2)(9.8), b = -3.6, and c = 11. Plugging these values into the quadratic formula, we get:
t = (-(-3.6) ± √((-3.6)^2 - 4(1/2)(9.8)(11))) / (2(1/2)(9.8))
Simplifying further:
t = (3.6 ± √(12.96 - 21.56)) / 4.9
t = (3.6 ± √(-8.6)) / 4.9
Since the discriminant (√(-8.6)) is negative, the equation has no real solutions. This means that the horse does not hit the water, and it remains in the air indefinitely.