a rock is throw upward with a velocity of 24 meters per second from the top of a 38 meter cliff, and it misses the cliff on the way back down. When will the rock be 2 meters from ground level?
the equation is
h = -4.9t^2 + 24t + 38 , where h is the height above ground level, and t is seconds
so you want
2 = -4.9t^2 + 24t + 38
4.9t^2 - 24t - 36 = 0
use the quadratic formula to find t, reject the negative value.
Well, well, well, looks like the rock is having quite the adventure. It's unlike any rock I've ever seen! Anyway, let's see if we can help it out.
So, the rock is thrown upward from a 38 meter cliff with an initial velocity of 24 m/s. Now, to find out when it will be 2 meters from ground level, we need to do a little bit of math. Don't worry, I'll do the calculating—I've got a degree in Rock-ology!
First things first, let's find out how high the rock goes before it starts coming down. We can use the good old kinematic equation:
v^2 = u^2 + 2as
Now, at the highest point, the rock's velocity would be zero. So, let's plug in the values we know:
0 = 24^2 + 2(-9.8)s
Solving this equation, we find that the rock reaches a maximum height of approximately 29.4 meters.
Next, let's figure out how long it takes for the rock to fall back down to a height of 2 meters. Again, we can use the same kinematic equation, this time solving for time:
s = ut + 0.5at^2
We know the initial height is 29.4 meters and the final height is 2 meters, so we can rearrange the equation to find time:
29.4 - 2 = 0 x t + 0.5 x (-9.8) x t^2
Simplifying further, we get:
t^2 - 2.99t - 5.88 = 0
Solving this quadratic equation, we find two possible values for time: t ≈ 0.59 seconds and t ≈ 9.91 seconds.
So, the rock will be 2 meters from the ground level approximately 0.59 seconds and 9.91 seconds after it's thrown. Keep an eye out for that adventurous rock!
To determine when the rock will be 2 meters from ground level, we can use the kinematic equation for vertical motion:
y = y0 + v0t - (1/2)gt^2
Where:
y = final displacement (2 meters)
y0 = initial displacement (38 meters)
v0 = initial velocity (24 meters per second)
g = acceleration due to gravity (-9.8 meters per second squared)
t = time
Let's solve this equation step-by-step:
1. Rearrange the equation to isolate t:
2 = 38 + (24)t - (1/2)(-9.8)t^2
2. Simplify the equation:
0 = -4.9t^2 + 24t + 36
3. Solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -4.9, b = 24, and c = 36.
t = (-24 ± √(24^2 - 4(-4.9)(36))) / (2(-4.9))
4. Evaluate the expression under the square root:
t = (-24 ± √(576 + 705.6)) / -9.8
t = (-24 ± √1281.6) / -9.8
5. Calculate the square root:
t = (-24 ± 35.8) / -9.8
6. Solve for t:
t1 = (-24 + 35.8) / -9.8
t1 ≈ 1.18 seconds
t2 = (-24 - 35.8) / -9.8
t2 ≈ 6.31 seconds
Therefore, the rock will be 2 meters from the ground level at approximately 1.18 seconds and 6.31 seconds after it is thrown upward.
To solve this problem, we can break it into two parts: the upward motion of the rock and the downward motion of the rock. We can use the laws of motion and the equations of kinematics to solve it step by step.
Step 1: Find the time it takes for the rock to reach its maximum height.
The initial velocity of the rock is 24 m/s, and the acceleration due to gravity is -9.8 m/s^2 (since the rock is moving upward, the acceleration is negative).
We can use the following kinematic equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
At the highest point, the final velocity is 0, so we have:
0 = 24 - 9.8t.
Solving for t, we get:
9.8t = 24,
t = 24 / 9.8 ≈ 2.45 seconds.
Therefore, it takes approximately 2.45 seconds for the rock to reach its maximum height.
Step 2: Find the time it takes for the rock to descend from its maximum height to the ground.
The initial height of the rock is 38 meters, and we want to find the time when it is 2 meters from the ground.
We can use the following kinematic equation:
s = ut + (1/2)at^2,
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values, we have:
-36 = 0t + (1/2)(-9.8)t^2,
-36 = -4.9t^2,
t^2 = 36 / 4.9 ≈ 7.35,
t ≈ √7.35 ≈ 2.71 seconds.
Therefore, it takes approximately 2.71 seconds for the rock to descend from its maximum height to 2 meters above the ground.
Step 3: Calculate the total time for the rock to reach a distance of 2 meters from the ground.
Since these two time intervals occur one after the other, we can simply add the times:
Total time = 2.45 + 2.71 ≈ 5.16 seconds.
Therefore, the rock will be 2 meters from the ground approximately 5.16 seconds after it is thrown.