A rock is thrown upward with a velocity of 25 meters per second from the top of a 50 meter high cliff, and it misses the cliff on the way back down. When will the rock be 7 meters from the water, below?
your height equation must be
h = -4.5t^2 + 25t + 50
so you want to solve:
7 = -4.5t^2 + 25t + 50
4.5t^2 - 25t - 43 = 0
use the quadratic formula, discard any negative value of t
To find the time when the rock will be 7 meters from the water below the cliff, we can use the equations of motion.
The equation to determine the height of the rock at a given time is:
h(t) = h0 + v0t - (1/2)gt^2
where:
h(t) is the height of the rock at time t
h0 is the initial height (from the top of the cliff)
v0 is the initial velocity
g is the acceleration due to gravity (-9.8 m/s^2)
Since the rock is thrown upward, the initial velocity v0 is positive (+25 m/s) and the initial height h0 is also positive (+50 m).
When the rock is at a height of 7 meters, we have:
h(t) = 7
h0 + v0t - (1/2)gt^2 = 7
Substituting the known values:
50 + 25t - (1/2) * 9.8 * t^2 = 7
Simplifying the equation:
-4.9t^2 + 25t + 43 = 0
Now we can solve this quadratic equation to find the time when the rock is 7 meters from the water below.
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where:
a = -4.9
b = 25
c = 43
Substituting the values into the quadratic formula:
t = (-25 ± sqrt(25^2 - 4 * -4.9 * 43)) / (2 * -4.9)
Simplifying the equation:
t = (-25 ± sqrt(625 + 843.2)) / (-9.8)
Calculating the square root:
t = (-25 ± sqrt(1468.2)) / (-9.8)
t ≈ (-25 ± 38.31) / (-9.8)
Now we have two possibilities for t:
1. t ≈ (-25 + 38.31) / (-9.8)
t ≈ 13.31 / (-9.8)
t ≈ -1.36 seconds
2. t ≈ (-25 - 38.31) / (-9.8)
t ≈ -63.31 / (-9.8)
t ≈ 6.46 seconds
Since time cannot be negative, we discard the negative value (-1.36 seconds).
Therefore, the rock will be 7 meters from the water below the cliff after approximately 6.46 seconds.
To determine when the rock will be 7 meters from the water below, we can use the kinematic equations of motion.
Step 1: Find the time it takes for the rock to reach its maximum height.
The initial velocity is 25 m/s, and the acceleration due to gravity is -9.8 m/s^2 (since it is acting in the opposite direction). The final velocity at the top of the trajectory will be 0 m/s. Using the kinematic equation:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (25 m/s)
a = acceleration (-9.8 m/s^2)
t = time
Substituting the known values:
0 = 25 - 9.8t
Rearranging the equation to solve for time, t:
9.8t = 25
t = 25 / 9.8
t ≈ 2.55 seconds
So, it takes approximately 2.55 seconds for the rock to reach its maximum height.
Step 2: Calculate the maximum height reached by the rock.
The equation to determine the displacement (h) of an object in free fall is:
h = ut + (1/2)at^2
where:
u = initial velocity (25 m/s)
a = acceleration (-9.8 m/s^2)
t = time taken to reach the maximum height (2.55 seconds)
Substituting the given values:
h = 25(2.55) + (1/2)(-9.8)(2.55)^2
h ≈ 31.76 meters
Therefore, the maximum height reached by the rock is approximately 31.76 meters.
Step 3: Find the time it takes for the rock to fall from the maximum height to a position 7 meters from the water below.
The total height from the top of the cliff to the water below is 50 meters, and the rock is 31.76 meters above the water at its maximum height. Therefore, the distance fallen from the maximum height to 7 meters above the water is:
31.76 - 7 = 24.76 meters
Now, we can use the equation for vertical displacement:
h = ut + (1/2)at^2
where:
h = vertical displacement (24.76 m)
u = initial velocity (0 m/s, as the rock is momentarily at rest at the maximum height)
a = acceleration (-9.8 m/s^2)
t = time taken to fall from maximum height to 7 meters above the water
Substituting the known values:
24.76 = 0(t) + (1/2)(-9.8)(t)^2
24.76 = -4.9t^2
Rearranging the equation:
t^2 = -24.76 / -4.9
t^2 ≈ 5.05
Taking the square root of both sides:
t ≈ √5.05
t ≈ ±2.25
Since time can't be negative, the rock will be at a position 7 meters from the water below approximately 2.25 seconds after reaching its maximum height.
Therefore, the rock will be 7 meters from the water below about 2.25 seconds after reaching its maximum height.