A stone is dropped from the top of a cliff which is 100m high. At the same time another stone is vertically thrown upwards from the foot of the cliff woth a speed of 40m/s.AT what height and when do the stones meet
See Related Questions: Thu, 4-21-11, 9:53 AM.
To find the height at which the two stones meet, we'll need to calculate the time it takes for each stone to reach that height.
Let's start with the stone that's dropped from the top of the cliff.
We can use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height
u = initial velocity (0 m/s since the stone is dropped)
t = time
g = acceleration due to gravity (-9.8 m/s^2, assuming standard gravity)
For the stone that's thrown upwards from the foot of the cliff, we'll need the time it takes for the stone to reach its maximum height. At that point, the stone starts falling, and the situation is similar to the stone that was dropped.
The equation of motion for this stone is:
v = u + gt
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (40 m/s)
t = time
g = acceleration due to gravity (-9.8 m/s^2)
First, let's calculate the time it takes for the stone thrown upwards to reach maximum height:
0 = 40 - 9.8t
9.8t = 40
t = 40 / 9.8
t ≈ 4.08 seconds
Now, we know it takes approximately 4.08 seconds for the stone thrown upwards to reach its maximum height.
Next, we'll calculate the height that the stone dropped from the top of the cliff reaches in the same amount of time.
h = 0 + (1/2)(-9.8)(4.08)^2
h ≈ -80.41 meters
Since distance cannot be negative in this context, we can ignore the negative sign, and the height is approximately 80.41 meters.
Therefore, the stones meet at a height of approximately 80.41 meters above the ground, after approximately 4.08 seconds.