George, a cliff diver, begins running 3.15 m away from the edge of a 33.3 m high cliff, if he accelerate at a constant rate of 0.76 m/s2 until he runs horizontally off the cliff, then how far away from the base will he hit the water? m

To find the distance George will hit the water, we need to find the time it takes for him to run off the cliff, and then use that time to find the horizontal distance he covers.

First, let's find the time it takes for George to run off the cliff:
Using the equation of motion s = ut + (1/2)at^2, where s is the displacement, u is the initial speed, a is the acceleration, and t is the time:

Since George is starting from rest (u = 0) and his displacement (s) is equal to the height of the cliff (33.3 m), we can rewrite the equation as:
33.3 = (1/2)(0.76)t^2

Simplifying the equation:
66.6 = 0.38t^2

Dividing both sides by 0.38:
t^2 = 66.6 / 0.38

t^2 ≈ 175.26

Taking the square root of both sides:
t ≈ √175.26

t ≈ 13.23 seconds (rounded to two decimal places)

Now that we have the time it takes for George to run off the cliff, we can find the horizontal distance he covers using the equation:
displacement = velocity × time

Since George accelerates horizontally, his horizontal velocity at the time he runs off the cliff (t = 13.23 s) will be:
velocity = acceleration × time
velocity = 0.76 m/s^2 × 13.23 s
velocity ≈ 10.05 m/s (rounded to two decimal places)

Finally, we can find the horizontal distance George covers using the formula:
distance = velocity × time
distance = 10.05 m/s × 13.23 s

distance ≈ 133.09 meters (rounded to two decimal places)

Therefore, George will hit the water approximately 133.09 meters away from the base of the cliff.

To solve this problem, we can break it down into two parts: the horizontal motion and the vertical motion of George.

Let's first look at the horizontal motion. George accelerates at a constant rate until he runs off the cliff. Since he runs horizontally, his final horizontal velocity will be constant throughout his motion. We can calculate his final horizontal velocity using the formula:

v = u + at

Where:
v = final velocity (which is the horizontal velocity)
u = initial velocity (which is 0 since George starts from rest)
a = acceleration (which is 0 since George is running horizontally off the cliff)
t = time

In this case, we need to find the time it takes for George to reach the edge of the cliff. We can use the following formula to calculate the time:

s = ut + (1/2)at^2

Where:
s = distance (which is the distance George runs before reaching the edge of the cliff)
u = initial velocity (which is 0 since George starts from rest)
a = acceleration (which is 0.76 m/s^2)
t = time

Rearranging the formula, we get:

t = sqrt((2s)/a)

Now, let's calculate the time it takes for George to reach the edge of the cliff:

s = 3.15 m (distance from the edge of the cliff)
a = 0.76 m/s^2

t = sqrt((2 * 3.15) / 0.76)
t = sqrt(6.3 / 0.76)
t = sqrt(8.289)

Therefore, t ≈ 2.88 seconds.

Now, let's move on to the vertical motion. George falls vertically from the edge of the cliff to the water below. We can calculate the time it takes for him to fall using the formula:

s = ut + (1/2)at^2

Where:
s = distance (which is the height of the cliff, 33.3 m)
u = initial velocity (which is 0 since George starts from rest)
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
t = time

Rearranging the formula, we get:

t = sqrt((2s)/a)

Now, let's calculate the time it takes for George to fall:

s = 33.3 m (height of the cliff)
a = 9.8 m/s^2 (acceleration due to gravity)

t = sqrt((2 * 33.3) / 9.8)
t = sqrt(66.6 / 9.8)
t = sqrt(6.7918)

Therefore, t ≈ 2.61 seconds.

Since the horizontal and vertical motions are happening simultaneously, the total time it takes for George to hit the water is the larger of the two times, which in this case is approximately 2.88 seconds.

Now, to find the horizontal distance George travels before hitting the water, we can use the formula:

s = vt

Where:
s = distance (which is what we need to find)
v = final velocity (which is the horizontal velocity)
t = time

From the horizontal motion, we found that the final horizontal velocity is the same as the initial horizontal velocity, which we calculated to be 3.67 m/s.

t = 2.88 seconds

s = (3.67 m/s)(2.88 s)
s ≈ 10.57 meters

Therefore, George will hit the water approximately 10.57 meters away from the base of the cliff.