A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 12.6 m higher than A, with speed v/2. Calculate the maximum height reached by the stone above point B.
I am still stuck on this problem, mainly because I can't seem to get v/2. I think I know all the steps...but not having v/2 is keeping me from doing them.
well, you know the amount of kinetic energy lost from a to b, and the change in Potential energy
1/2 mv^-1/2 m (v/2)^2= mg12.5
3/4 v^2=12.5*9.8
solve for v.
To solve this problem, we need to use the principles of projectile motion. Let's break it down step by step:
1. Identify the given information:
- The stone is thrown vertically upward.
- It passes point A with speed v.
- Point B is 12.6 m higher than point A, and the stone passes it with a speed of v/2.
2. Recall the equations of motion for vertical motion under constant acceleration:
- v = u + gt (equation 1)
- s = ut + (1/2)gt^2 (equation 2)
- v^2 = u^2 + 2gs (equation 3)
3. Since the stone is thrown vertically upward, its initial velocity (u) will be positive, and acceleration due to gravity (g) will be negative (-9.8 m/s^2).
4. At point A:
- We know the initial velocity (u) is v.
- We don't know the time (t) or the displacement (s).
From equation 1:
v = u + gt
v = v + (-9.8)t
0 = -9.8t
This implies that at point A, the stone reaches the highest point and momentarily stops before coming back down. Hence, t = 0.
5. At point B:
- We know the initial velocity (u) is v/2.
- We don't know the time (t) or the displacement (s).
Using equation 1:
(v/2) = (v/2) + (-9.8)t
0 = (-9.8)t
Again, this implies that at point B, the stone momentarily stops before changing direction. Hence, t = 0.
6. To find the maximum height reached by the stone above point B, we need to determine the displacement (s) from B.
Using equation 2:
s = ut + (1/2)gt^2
s = ((v/2)(0)) + (1/2)(-9.8)(0)^2
s = 0
Therefore, at the maximum height above point B, the displacement (s) is zero. This means that the stone reaches its highest point above point B and then falls back down.
To summarize, the stone does not reach any maximum height above point B.