A rock is thrown upward with a velocity of 15 meters per second from the top of a 30 meter high cliff, and it misses the cliff on the way back down. When will the rock be 10 meters from the water, below? Round your answer to two decimal places.
I am not understanding how to start this
4.07 seconds
Better review your motion equation.
h = Ho + Vo*t - 1/2 gt^2
the height above the water, using g = 10, is thus
h = 30 + 15t - 5t^2
So, just solve for t when h=10
A rock is thrown upward with a velocity of 15 meters per second from the top of a 34 meter high cliff, and it misses the cliff on the way back down. When will the rock be 4 meters from ground level? Round your answer to two decimal places.
Well, it sounds like the rock is causing quite the splash! Let's break this problem down. We have a rock that is thrown upward with a velocity of 15 meters per second from a height of 30 meters. We want to find out when the rock will be 10 meters from the water below.
To solve this, we need to use some basic kinematics equations. In this case, because the rock is thrown upward, we'll use the equation:
h = h0 + v0t - 0.5gt^2
Where:
h is the height of the rock at any given time
h0 is the initial height (30 meters in this case)
v0 is the initial velocity (15 meters per second in this case)
t is the time in seconds
g is the acceleration due to gravity (approximately 9.8 m/s^2)
We want to find out when the rock will be 10 meters from the water, so we can set h equal to 10 and solve for t.
10 = 30 + 15t - 0.5(9.8)t^2
Simplifying this equation will give us a quadratic equation, which we can solve for t. Once we find the two values of t, we can disregard the negative solution as we are only interested in when the rock is on its way back down.
So, let's crunch those numbers and see when the rock will be making a big splash!
To solve this problem, you can consider the motion of the rock in two separate phases: when it is moving upward, and when it is moving downward.
First, let's address the upward motion of the rock. We need to find the time taken by the rock to reach its maximum height (the highest point). We can use the equation of motion:
v = u + at
Where:
- v is the final velocity (0 m/s in this case, as the rock comes to rest at the highest point)
- u is the initial velocity (15 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2, taking downward direction as negative)
- t is the time taken
At the highest point, the velocity is zero, so we can rearrange the equation to solve for time:
0 = 15 + (-9.8)t
Simplifying the equation gives:
9.8t = 15
Now, we can solve for t:
t = 15 / 9.8
Calculating this, t is approximately 1.53 seconds.
Next, let's consider the downward motion of the rock. We need to find out when the rock will be 10 meters from the water below. We can use the equation of motion again:
s = ut + (1/2)at^2
Where:
- s is the displacement (10 meters below the top of the cliff)
- u is the initial velocity (0 m/s, as the rock is momentarily at rest at the highest point)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken (which we already found to be approximately 1.53 seconds)
Substituting the values into the equation, we get:
10 = 0 + (1/2)(-9.8)(1.53)^2
Simplifying the equation gives:
10 = -7.598t^2
Rearranging the equation, we need to solve for t:
t^2 = -10 / -7.598
Taking the square root of both sides, we get:
t = sqrt(10 / 7.598)
Evaluating this expression, t is approximately 1.28 seconds.
Therefore, the rock will be 10 meters from the water below after approximately 1.28 seconds.