A rock is thrown upward with a velocity of 28 meters per second from the top of a 38 meter high cliff, and it misses the cliff on the way back down. When will the rock be 7 meters from the water, below? Round your answer to two decimal places.
Gravity Formula
6.46
Ah, the gravity formula! It's like the secret sauce for falling objects. Let me crunch some numbers for you, but don't worry, I promise to keep it light and entertaining!
First things first, let's divide this problem into two parts: when the rock is going up and when it's coming back down. We have the initial velocity (28 m/s), the height of the cliff (38 meters), and we want to find out when the rock will be 7 meters from the water, below.
During the upward journey, the velocity will decrease until it reaches zero at the highest point. At that point, the only force acting on the rock will be gravity, bringing it back down.
Okay, let's do some calculations. Time to channel my inner mathematician!
Using the equation for vertical motion under gravity, the time it takes for the rock to go up and reach its highest point can be calculated with the formula:
t = (vf - vi) / g
Where:
- vf is the final velocity (which we already know is 0 at the highest point).
- vi is the initial velocity (28 m/s in this case).
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now let's plug in the values and solve for t:
t = (0 - 28) / -9.8
t = 2.857...
So, it takes approximately 2.86 seconds for the rock to reach its highest point.
The next step is to calculate how long it will take for the rock to fall from the highest point to 7 meters below the water surface. The time it takes to fall is equal to the time it took to go up, so we have:
t_total = 2*t
t_total = 2 * 2.86
t_total = 5.72 seconds
So, the rock will be 7 meters from the water, below at approximately 5.72 seconds. Remember to keep that round!
And there you have it! The answer to your question, served with a side of humor. If you have any more questions or need further assistance, feel free to ask!
The gravity formula is given by:
h = ut + (1/2)gt²
where:
h = height of the object at any given time (in this case, the height from the top of the cliff to the water)
u = initial velocity (in this case, the velocity at which the rock is thrown upward)
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time
In this case, we want to find the time when the rock is 7 meters from the water below. So, we need to rearrange the formula to solve for time (t):
7 = -38 + 28t + (1/2)(-9.8)t²
Simplifying this equation will give us a quadratic equation:
4.9t² + 28t - 31 = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
where:
a = 4.9
b = 28
c = -31
Applying these values, we get:
t = (-28 ± √(28² - 4 * 4.9 * -31)) / (2 * 4.9)
Simplifying further will give us two values for time (t). Let's calculate them:
t₁ = (-28 + √(28² - 4 * 4.9 * -31)) / (2 * 4.9)
t₂ = (-28 - √(28² - 4 * 4.9 * -31)) / (2 * 4.9)
t₁ ≈ 1.95 seconds
t₂ ≈ -3.18 seconds
Since time cannot be negative in this case, we can ignore the negative value. Therefore, the rock will be 7 meters from the water below after approximately 1.95 seconds.
6.64 goofy.
from your data, if h is the height above the water
h = -4.9t^2 + 28t + 38
set h = 7
4.9t^2 - 28t - 31 = 0
use the quadratic formula to solve for t, use the positive result