A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is thrown vertically upwards with a velocity of 25 m/s. When and where will the two stones cross each other?
when:
100-4.9t^2 = 25t-4.9t^2
where: evaluate either expression at the found value of t.
To find when and where the two stones will cross each other, we need to consider the motion of each stone separately.
First, let's determine the time it takes for the stone thrown upwards to reach its highest point (the point where its velocity becomes zero) and start falling back down.
Using the kinematic equation for vertical motion:
v = u + at
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (25 m/s)
a = acceleration (-9.8 m/s^2 due to gravity)
t = time
Substituting the values:
0 = 25 - 9.8t
Solving the equation for t:
9.8t = 25
t ≈ 2.55 seconds
So, it will take approximately 2.55 seconds for the stone thrown upwards to reach its highest point.
Next, let's find the time it takes for the stone falling downwards to reach the same height (100 m).
Using the kinematic equation:
s = ut + (1/2)at^2
where:
s = distance (100 m)
u = initial velocity (0 m/s)
a = acceleration (-9.8 m/s^2 due to gravity)
t = time
Substituting the values:
100 = 0 + (1/2)(-9.8)t^2
Simplifying the equation:
-4.9t^2 = -100
Dividing both sides by -4.9:
t^2 ≈ 20.41
Taking the square root of both sides:
t ≈ 4.52 seconds
So, it will take approximately 4.52 seconds for the stone falling downwards to reach the height of 100 m.
Since the stone thrown upwards reaches its highest point in 2.55 seconds and the stone falling downwards takes 4.52 seconds to reach a height of 100 m, they will cross each other somewhere in between.
To determine the exact height where they cross, we can use the equation:
s = ut + (1/2)at^2
Substituting the values for the stone thrown upwards:
s = 25(2.55) + (1/2)(-9.8)(2.55)^2
Solving the equation:
s ≈ 32.12 meters
Therefore, the two stones will cross each other approximately 32.12 meters above the ground.