A rock is thrown upward with a velocity of 15 meters per second from the top of a 26 meter high cliff ad i misses the cliff on the way back down. when will the rock be 5 meters from the water below?
height = -4.9t^2 + 15t + 26
When the rock is 5 metres above the water,
5 = -4.9t^2 + 15t + 26
4.9t^2 - 15t - 21 = 0
t = (15 ± √636.6)/9.8
= 4.1052 seconds or a negative
After appr 4.1 seconds, the rock is 5 metres above the water.
To solve this problem, we can use the equations of motion to find the time when the rock is 5 meters above the water below.
We know that the initial velocity of the rock when it is thrown upward is 15 meters per second, and the height of the cliff is 26 meters.
First, let's find the time it takes for the rock to reach its maximum height. We can use the equation:
v = u + at
where:
v = final velocity at maximum height (0 m/s because the rock reaches its highest point and then starts falling back down)
u = initial velocity (15 m/s)
a = acceleration due to gravity (-9.8 m/s^2, taking downward direction as negative)
t = time
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Substituting the values, we get:
t = (0 - 15) / -9.8
t = 1.53 seconds (approximately)
So, it takes 1.53 seconds for the rock to reach its maximum height.
Next, let's determine the total time of flight (time taken for the rock to go up and down). The total time of flight is double the time it took to reach the maximum height.
Therefore, the total time of flight is 2 * 1.53 = 3.06 seconds.
Now let's find the position of the rock after 3.06 seconds. We can use the equation:
s = ut + (1/2)at^2
where:
s = distance traveled (unknown, 5 meters in this case)
u = initial velocity (15 m/s)
t = time (3.06 seconds)
a = acceleration due to gravity (-9.8 m/s^2)
Rearranging the equation to solve for distance (s), we have:
s = ut + (1/2)at^2
Substituting the values, we get:
5 = (15 * 3.06) + (1/2)(-9.8)(3.06^2)
Simplifying the equation, we find:
5 = 45.9 - 45.5
5 = 0.4
This is not possible since 5 is not equal to 0.4. Therefore, the rock does not come 5 meters from the water below.
Hence, the rock will not be 5 meters from the water below at any point during its trajectory.