A rock is thrown upward with a velocity of 28 meters per second from the top of a 36 meter high cliff, and it misses the cliff on the way back down. When will the rock be 9 meters from the water below. Round to two decimal places.
we have
h(t) = 36 + 28t - 4.9t^2
So, when h=9, just solve
36 + 28t - 4.9t^2 = 9
t ~= 6.55 sec
To solve this problem, we need to find the time it takes for the rock to reach a height of 9 meters while moving upward and downward.
Let's start by finding the time it takes for the rock to reach its highest point, which is when its velocity becomes 0 m/s. We can use the formula for vertical motion:
v = u + at
Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (28 m/s)
a = acceleration due to gravity (-9.8 m/s^2, taking negative since it opposes the upward motion)
t = time
Plugging in the values, we get:
0 = 28 + (-9.8)t
Rearranging the equation, we have:
9.8t = 28
t = 28 / 9.8
t ≈ 2.86 seconds
So, it takes approximately 2.86 seconds for the rock to reach its highest point.
Next, we can find the time it takes for the rock to fall from the top of the cliff to a height of 9 meters. We can use the formula:
s = ut + (1/2)at^2
Where:
s = displacement (36 - 9 = 27 meters, taking negative since it is downward)
u = initial velocity (0 m/s at the highest point)
a = acceleration due to gravity (9.8 m/s^2, taking positive since it is downward)
t = time
Plugging in the values, we get:
-27 = 0 + (1/2)(9.8)t^2
Rearranging the equation, we have:
4.9t^2 = 27
t^2 = 27 / 4.9
t ≈ √(5.51)
t ≈ 2.35 seconds
So, it takes approximately 2.35 seconds for the rock to fall from the top of the cliff to a height of 9 meters.
Since we are looking for the total time, we need to add the time it takes for the rock to reach its highest point and the time it takes for the rock to fall from the top of the cliff to a height of 9 meters:
Total time = 2.86 + 2.35
Total time ≈ 5.21 seconds
Therefore, the rock will be approximately 9 meters from the water below after 5.21 seconds.