A stone with a weight of 5.38 N is launched vertically from ground level with an initial speed of 25.0 m/s, and the air drag on it is 0.269 N throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground
To find the maximum height reached by the stone, we need to determine when the stone reaches its highest point. At this point, the stone's final velocity will be zero. We can use the kinematic equation for vertical motion to solve for the maximum height.
The equation for vertical motion without considering air resistance is:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
In this case, we are looking for the maximum height reached by the stone, so v = 0 and a = -9.8 m/s^2 (acceleration due to gravity).
Using the equation, we can rearrange it to solve for s:
s = (v^2 - u^2) / (2a)
Substituting the given values:
u = 25.0 m/s
v = 0
a = -9.8 m/s^2
s = (0^2 - (25.0 m/s)^2) / (2 * -9.8 m/s^2)
s = -625.0 m^2/s^2 / -19.6 m/s^2
s = 31.89 m
Therefore, the maximum height reached by the stone is 31.89 meters.
Now, let's calculate the speed of the stone just before it hits the ground. To do this, we'll need to consider the effect of air drag.
The net force acting on the stone can be calculated using Newton's second law:
F_net = mg - F_drag
where:
F_net = net force
m = mass
g = acceleration due to gravity
F_drag = drag force
We can rearrange the equation to solve for the mass of the stone:
m = (F_net + F_drag) / g
Substituting the given values:
F_net = 5.38 N
F_drag = 0.269 N
g = 9.8 m/s^2
m = (5.38 N + 0.269 N) / 9.8 m/s^2
m = 5.649 kg
Now, we can calculate the acceleration of the stone:
a = (F_net - F_drag) / m
a = (5.38 N - 0.269 N) / 5.649 kg
a = 0.945 m/s^2
Using the kinematic equation, we can find the final velocity of the stone just before it hits the ground:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
In this case, u = 0 (since the stone is dropped), a = 0.945 m/s^2, and s = -31.89 m (negative because the stone is moving downward).
v^2 = 0^2 + 2 * 0.945 m/s^2 * (-31.89 m)
v^2 = -60.072 m^2/s^2
v = -7.75 m/s (since we are interested in magnitude, the negative sign can be ignored)
Therefore, the speed of the stone just before it hits the ground is 7.75 m/s.
To find the maximum height reached by the stone, we need to determine the point at which the stone's velocity becomes zero.
We can find this point by using the equation of motion: v^2 = u^2 + 2as, where:
- v is the final velocity (zero in this case),
- u is the initial velocity (25.0 m/s),
- a is the acceleration, and
- s is the displacement.
The acceleration can be calculated using Newton's second law: F = ma, where F is the net force acting on the stone, and m is its mass.
First, let's determine the net force acting on the stone. In this case, the net force is the difference between the weight of the stone and the air drag force:
Net force = Weight - Air drag
The weight of the stone can be determined using the formula: Weight = mass * gravitational acceleration.
Given that the weight of the stone is 5.38 N, we have:
5.38 N = m * 9.8 m/s^2
Solving this equation for mass, we find the mass of the stone as follows:
m = 5.38 N / 9.8 m/s^2
Next, we can calculate the net force:
Net force = 5.38 N - 0.269 N
Now, using Newton's second law, we can determine the acceleration:
Net force = m * a
5.111 N = (5.38 N / 9.8 m/s^2) * a
Solving for acceleration, we have:
a = 5.111 N * (9.8 m/s^2 / 5.38 N)
Now, substituting the values into the equation of motion: v^2 = u^2 + 2as
0 = (25.0 m/s)^2 + 2 * a * s
Since we are interested in finding the maximum height, we know that at the maximum height, the final velocity (v) is zero. Thus, the equation becomes:
0 = (25.0 m/s)^2 + 2 * a * s_max
We can now solve for the maximum height (s_max).