A rock is thrown upward with a velocity of 12 meters per sec from the top of a 25 meter high cliff and it misses on it's way back down. When will the rock be 3 meters from water
3.67
Well, that's quite the puzzler! I know just the thing to lighten the mood. Why did the rock go to therapy? Because it had trouble letting go!
Now, back to your question. Let's break it down. Since the rock is thrown upward, we know it will reach its highest point and start coming back down. To figure out when it will be 3 meters from the water, we need to find the time it takes for the rock to reach that height.
We can use the equations of motion to solve this. The equation we need is:
h = v0t - (1/2)gt^2
where h is the height above the water, v0 is the initial velocity (12 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Let's set h to 3 meters and solve for t:
3 = 12t - (1/2)(9.8)t^2
To make it easier, we can multiply everything by 2 to eliminate the fraction:
6 = 24t - 9.8t^2
Now we have a quadratic equation! We can rearrange it to:
9.8t^2 - 24t + 6 = 0
Solving this equation will give us the time at which the rock will be 3 meters from the water. Bring on the math wizards!
To determine when the rock will be 3 meters from the water, we need to find the time it takes for the rock to reach that height. Let's break down the problem into smaller parts to solve it step by step.
Step 1: Calculate the time it takes for the rock to reach its maximum height.
The initial velocity of the rock when thrown upward is 12 meters per second. The acceleration due to gravity is approximately 9.8 meters per second squared. Since the rock is thrown vertically, the only force acting on it is gravity, which causes it to decelerate.
We can use the following equation to calculate the time it takes for the rock to reach its maximum height:
v = u + at
Here:
v = final velocity (0 m/s at the topmost point since the rock comes to rest momentarily)
u = initial velocity (12 m/s when thrown upward)
a = acceleration due to gravity (-9.8 m/s², negative because it acts in the opposite direction to motion)
t = time taken (unknown)
Using the equation, we can rearrange it to solve for t:
t = (v - u) / a
Substituting the given values:
t = (0 - 12) / -9.8
t ≈ 1.224 seconds
So, it takes approximately 1.224 seconds for the rock to reach its maximum height.
Step 2: Calculate the total time the rock is in the air.
Since the rock comes back down to the same height it was thrown from, the total time it spends in the air is twice the time it takes to reach the maximum height.
Total time in air = 2 * t
Total time in air = 2 * 1.224
Total time in air ≈ 2.448 seconds
So, the rock is in the air for approximately 2.448 seconds.
Step 3: Calculate the distance the rock falls during the given time.
To find when the rock will be 3 meters from the water surface, we need to calculate the distance the rock falls during the total time it is in the air.
Using the kinematic equation: s = ut + (1/2)at²
Here:
s = distance (unknown)
u = initial velocity (0 m/s)
t = time (2.448 seconds)
a = acceleration due to gravity (-9.8 m/s²)
Rearranging the equation:
s = (1/2)at²
Substituting the values:
s = (1/2) * (-9.8) * (2.448)²
s ≈ -30.03 meters
The distance the rock falls during the given time is approximately 30.03 meters.
Step 4: Calculate the height from the top of the cliff to the water's surface.
The cliff's height is given as 25 meters. Since the rock is thrown upwards from the top of the cliff, the total distance from the water's surface to the top of the cliff would be 25 - 30.03 meters (negative sign as the rock falls down).
Height from top of the cliff to the water's surface = 25 - 30.03
Height from top of the cliff to the water's surface ≈ -5.03 meters
The height from the top of the cliff to the water's surface is approximately -5.03 meters.
Since the calculated height is negative, it means the rock will already be below the water's surface before it reaches 3 meters from it. Therefore, the rock will not reach a height of 3 meters from the water.
find t when
25 + 12t - 5t^2 = 3
using g=10 (easier than using 9.81)