A stone is projected vertically upwards with a velocity of 20m/s. 2s later another stone is similarly projecte?
with the same velocity. when the two stones meet the second one is rising at a velocity of 10m/s.neglecting air resistance calculate the length of time the second stone is in motion before they met.
the velocity of the first stone when they met.take gravity to be 10m/s.
Stone 1
h = 0 + 20 t - 5 t^2
v = 20 - 10 t
Stone 2
h = 0 + 20(t-2) - 5 (t^2 - 4 t + 4)
v = 20 - 10(t-2)
==========================
at h, v of stone 2 = +10
10 = 20 -10 t + 20
10 t = 30
t = 3 seconds that the first one is in the air
t-2 = 1 second for the second one (first question)
v = 20 - 10*3 = -10 m/s for the first stone at collision
Stone #1:
V = Vo + g*t = 20 + (-10)*2 = 0 m/s.(h = h max.).
Stone #2:
a. V = Vo + g*t = 10.
20 + (-10)t = 10,
t = 1 s.
Stone #1:
b. V = Vo + g*t = 0 + 10*(3-2) = 10 m/s. = Velocity of 1st. stone when they
met.
Note: The 1st. stone had been in air 3 s(1+2) when they met. But 2 of those
3 s was rise time, That left 1 s fall time.
Well, it seems like these stones are having quite the risqué meeting! Let's calculate their encounter in the most humorous way possible, shall we?
First, let's find out how high the first stone goes. We know it's projected vertically upwards with a velocity of 20m/s and gravity is 10m/s. Using the formula v = u + gt (where v is final velocity, u is initial velocity, g is gravity, and t is time), we get:
0 = 20 - 10t1
10t1 = 20
t1 = 2 seconds
So, the first stone takes 2 seconds to reach its highest point.
Now, let's find out how high the second stone has risen in these 2 seconds. Again, using the formula v = u + gt, we know that the final velocity of the second stone is 10m/s, the initial velocity is also 20m/s (same as the first stone), and gravity is still 10m/s:
10 = 20 - 10t2
10t2 = 10
t2 = 1 second
Ah, so it seems the second stone rises for only 1 second before they meet! They must have quite the sense of urgency.
Now, let's find out the velocity of the first stone when they meet. We can use the formula v = u + gt, where v is the final velocity (which we don't know), u is the initial velocity (20m/s), g is gravity (10m/s), and t is the time they both meet (which we don't know yet).
v = 20 + 10t
But we know that the second stone reaches a velocity of 10m/s at that time, so we have:
10 = 20 + 10t
10t = -10
t = -1 second
Wait a minute, a negative time? That's impossible! Seems like these stones are stuck in some kind of time warp or have mastered time travel. In any case, we can't have a negative time, so it seems these stones will never meet in reality.
That's the thing about physics - sometimes it just can't keep up with the weirdness of reality!
To solve this problem, we need to find the time it takes for the first stone to reach its highest point, and then find the distance it covers during that time.
Let's start by finding the time it takes for the first stone to reach its highest point. We know that the initial velocity of the first stone is 20 m/s, and that the acceleration due to gravity is 10 m/s^2 (assuming downwards direction is positive).
We can use the equation for vertical motion:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. At the highest point, the final velocity becomes zero, so we have:
0 = 20 - 10t1
Simplifying this equation, we find:
10t1 = 20
t1 = 2 seconds
Therefore, it takes the first stone 2 seconds to reach its highest point.
Now we can find the distance covered by the first stone during this time. We can use the equation for displacement during vertical motion:
d = vit + (1/2)at^2
where d is the displacement, vi is the initial velocity, a is the acceleration, and t is the time. At the highest point, the displacement becomes zero, so we have:
0 = 20(2) + (1/2)(-10)(2)^2
Simplifying this equation, we find:
0 = 40 - 20
Therefore, the first stone covers a distance of 40 meters during the 2 seconds it takes to reach its highest point.
Since the second stone is launched with the same velocity, it will also take 2 seconds to reach the same height. However, since the second stone is launched 2 seconds after the first stone, it will be in motion for 2 seconds less than the first stone.
Therefore, the second stone is in motion for:
2 seconds - 2 seconds = 0 seconds
So the second stone is not in motion when they meet.
As for the velocity of the first stone when they meet, it will be equal to the initial velocity of the first stone because it is still moving upwards. Therefore, the velocity of the first stone when they meet is 20 m/s.