A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 7.85 m higher than A, with speed v/2. Calculate (a) the speed v and (b) the maximum height reached by the stone above point B.
To solve this problem, we can use the principles of kinematics and the equations of motion. Let's break down the problem into two parts.
Part (a): Calculating the speed v
At point A, the stone has a speed v. We know that as the stone reaches point B, which is 7.85 m higher than A, its speed reduces to v/2. We also know that when the stone is at the maximum height, its speed will be zero.
Let's use the equations of motion to solve for the speed v at point A:
Using the equation of motion for vertical motion:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
At point A:
Initial velocity, u = v,
Final velocity, v = v/2,
Displacement, s = 7.85 m,
Acceleration, a = -9.8 m/s^2 (assuming upward direction as positive).
Plugging these values into the equation of motion, we get:
(v/2)^2 = v^2 + 2(-9.8)(7.85).
Simplifying the equation:
v^2/4 = v^2 - (2)(-9.8)(7.85).
Expanding and rearranging the equation:
(v^2 - 4v^2/4) = 2(9.8)(7.85),
Simplifying further:
-3v^2/4 = 2(9.8)(7.85).
After canceling out common terms:
-3v^2/4 = 153.18.
Multiplying both sides by -4/3:
v^2 = -204.24.
Taking the square root of both sides:
v ≈ √(-204.24) (ignoring the negative value since velocity cannot be negative),
v ≈ √204.24,
v ≈ 14.29 m/s.
Therefore, the speed of the stone at point A is approximately 14.29 m/s.
Part (b): Calculating the maximum height reached by the stone above point B
To calculate the maximum height reached by the stone above point B, we can use the formula for the maximum height of an object in vertical motion:
h_max = (v^2)/(2g),
where h_max is the maximum height, v is the final velocity (which is zero at the maximum height), and g is the acceleration due to gravity.
Plugging in the values:
v = 0 (since the velocity is zero at the maximum height),
g = -9.8 m/s^2 (assuming upward direction as positive).
h_max = (0^2)/(2(-9.8)),
h_max = 0 m.
Therefore, the maximum height reached by the stone above point B is 0 m, indicating that it does not go any higher than point B.
vf^2=Vi^2 +2gh
I assume you memorized that equation. Here we have.
(v/2)^2=v^2-2*g*7.85
solve for v.
Now, knowing the v, use the same equation as you started with (now Vf=0 at top, and vi=V/2, and you solve for h.