A rock is thrown upward with a velocity of 20 meters per second from the top of a 37 meter high cliff, and it misses the cliff on the way back down. When will the rock be 44 meters from the water, below? Round your answer to two decimal places.
To find the time when the rock is 44 meters from the water below, we can use the equation of motion for objects in free fall:
h = h0 + v0t - (1/2)gt^2
where:
h is the height of the object at time t
h0 is the initial height of the object
v0 is the initial velocity of the object
g is the acceleration due to gravity (approximated as 9.8 m/s^2)
t is the time elapsed
Let's solve the equation for the given values:
h = 44 meters
h0 = 37 meters
v0 = 20 m/s
g = 9.8 m/s^2
Substituting these values into the equation, we have:
44 = 37 + 20t - (1/2)(9.8)(t^2)
To solve for t, let's rearrange the equation and set it equal to zero:
0 = -4.9t^2 + 20t - 7
Now we can solve this quadratic equation. Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac))/(2a)
where:
a = -4.9
b = 20
c = -7
Substituting these values into the formula, we have:
t = (-20 ± sqrt(20^2 - 4(-4.9)(-7)))/(2(-4.9))
Simplifying the equation further:
t = (-20 ± sqrt(400 - 137.2))/(-9.8)
t = (-20 ± sqrt(262.8))/(-9.8)
Now, we can use a calculator to evaluate the square root:
t = (-20 ± 16.21)/(-9.8)
Now we can solve for t using both the positive and negative square root:
t1 = (-20 + 16.21)/(-9.8)
t2 = (-20 - 16.21)/(-9.8)
Calculating these values:
t1 ≈ -0.37 seconds
t2 ≈ 3.49 seconds
Since time cannot be negative in this context, we can ignore the negative value. Therefore, the rock will be 44 meters from the water below after approximately 3.49 seconds.
To determine when the rock will be 44 meters from the water below, we can use the principles of projectile motion and solve for time.
First, let's define the relevant variables:
- Initial velocity (u) = 20 meters per second (since the rock is thrown upward)
- Initial height (h) = 37 meters (height of the cliff)
- Final height (h') = -44 meters (below the water's surface, so negative)
- Acceleration due to gravity (g) = -9.8 meters per second squared (negative because it acts in the opposite direction of motion)
Now, we can use the formula for calculating the height (h) of an object in projectile motion at any given time (t):
h = h0 + ut + (1/2) * gt^2
We know that when the rock is 44 meters below the water's surface, the height (h) is -44 meters. Plugging in the values:
-44 = 37 + 20t - (1/2)(9.8)t^2
Simplifying the equation:
-44 - 37 = 20t - (1/2)(9.8)t^2
-81 = 20t - 4.9t^2
Rearranging the equation into a quadratic form:
4.9t^2 - 20t - 81 = 0
Now, we can solve this quadratic equation for t. We can do this by factoring, using the quadratic formula, or graphing. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 4.9, b = -20, and c = -81. Substituting these values:
t = (-(-20) ± √((-20)^2 - 4(4.9)(-81))) / (2(4.9))
Simplifying the equation:
t = (20 ± √(400 + 1581.6)) / 9.8
t = (20 ± √1981.6) / 9.8
Now, we can find the two values of t:
t1 = (20 + √1981.6) / 9.8
t2 = (20 - √1981.6) / 9.8
Calculating these values using a calculator:
t1 ≈ 5.49 seconds
t2 ≈ -2.05 seconds
Since time cannot be negative in this context, we can discard the negative value. Therefore, the rock will be 44 meters from the water below approximately 5.49 seconds after it is thrown upward from the top of the cliff.
h = 1/2 g t^2 + v t + Ho
44 = -4.9 t^2 + 20 t + 37
solve the quadratic for t