A rock is thrown upward with a velocity of 24 meters per second from the top of a 50 meter high cliff and it misses the cliff on the way back down. When will the rock be 7 meters from the water below
Well, well, well, it seems like this rock has some serious airtime ambitions! Let's break it down, shall we?
The rock is thrown upward, so we know gravity will eventually bring it back down. Now, if we imagine the rock's motion as a lovely parabolic arc, it will reach a certain height before it starts its descent.
Since the cliff is 50 meters high and the rock misses it on the way back down, it means the highest point the rock reaches is 50 meters. We can consider this as the peak of its journey.
Now, to find out when the rock will be 7 meters from the water below, we need to figure out how long it takes for the rock to reach that height while falling downward.
We can use a little physics to crack this one. The equation we'll employ is:
h = (initial velocity * time) - (0.5 * gravity * time^2),
where:
h is the distance traveled vertically,
initial velocity is the velocity at the start,
gravity is about 9.8 m/s^2 (let's ignore air resistance for simplicity), and
time is the time it takes to reach the desired height.
In this case, h = 7 meters, the initial velocity is the same 24 meters per second, and gravity, as always, is a reliable 9.8 m/s^2.
We can plug in these numbers and solve for time. After a little math magic, we find that the time it takes for the rock to be 7 meters from the water is approximately 0.77 seconds.
So, my friend, after all the calculations and the comedy show, the rock will be 7 meters above the water in about 0.77 seconds. Remember, don't get too close when it lands—rocks aren't known for their gentle dismounts!
To find out when the rock will be 7 meters from the water below, we need to determine the time it takes for the rock to reach its maximum height when thrown upward and the time it takes to fall from that maximum height to a height of 7 meters.
Step 1: Calculate the time to reach the maximum height.
Using the formula for the time it takes to reach maximum height in a vertical projectile motion:
t = (v - u) / g
where:
t = time
v = final velocity (0 m/s at maximum height)
u = initial velocity (24 m/s)
g = acceleration due to gravity (-9.8 m/s^2, taking the negative value as the object is thrown upward)
Substituting the given values into the equation,
t = (0 - 24) / -9.8
t = 24 / 9.8
t ≈ 2.45 seconds
Step 2: Calculate the time to fall from maximum height to 7 meters.
We can use the equation for vertical displacement of an object in free fall:
s = ut + (1/2)gt^2
Rearranging the equation for time, we get:
t = sqrt(2s/g)
where:
t = time
s = vertical displacement (from maximum height to 7 meters, 50 - 7 = 43 meters)
g = acceleration due to gravity (-9.8 m/s^2)
Substituting the given values into the equation,
t = sqrt(2 * 43 / -9.8)
t = sqrt(86 / -9.8)
t ≈ 3.07 seconds
Step 3: Add the times taken in the two steps.
The total time taken for the rock to be 7 meters from the water below is the sum of the times calculated in Steps 1 and 2:
Total time = 2.45 + 3.07
Total time ≈ 5.52 seconds
Therefore, the rock will be approximately 7 meters from the water below after approximately 5.52 seconds.
To find the time when the rock will be 7 meters from the water below, we can use the equations of motion. First, let's consider the upward motion of the rock.
We can use the equation:
h = ut + (1/2)gt^2
Where:
h = height of the rock at any given time
u = initial velocity (upward) = 24 m/s (positive because it is upward)
g = acceleration due to gravity = -9.8 m/s^2 (negative because it is acting downward)
t = time
Since the rock starts from the top of a 50 meter high cliff, the initial height would be 50 meters. We need to find the time when the height of the rock will be 7 meters.
Using the equation, we can rewrite it as:
7 = 24t - (1/2)(9.8)t^2
Rearranging the equation, we get:
0 = (1/2)(9.8)t^2 - 24t + (50 - 7)
Simplifying further:
0 = 4.9t^2 - 24t + 43
This equation is a quadratic equation in the form of at^2 + bt + c = 0. We can solve it by factoring or by using the quadratic formula.
By factoring, we can find that:
0 = (4.9t - 7)(t - 6)
This gives us two possible solutions for t:
1) 4.9t - 7 = 0 → 4.9t = 7 → t ≈ 1.43 seconds
2) t - 6 = 0 → t = 6 seconds
Since the rock starts its upward journey from the top of the cliff, we can discard the solution t = 6 seconds as it corresponds to when the rock reaches the top of its trajectory again. Therefore, the rock will be 7 meters from the water below approximately 1.43 seconds after it was thrown upward.
just solve for t in
50 + 24t - 4.9t^2 = -43