a ball with mass 0.15 kg is thrown upward with initial velocity 20 m/sec from the roof of a building 30 m high. there is a force due to air resistance of |v|/30, where velocity v is measured in m/sec.
a. find the maximum height above the ground the ball reaches.
b. find the time the ball hits the ground.
you cannot use the kinematic equations.
This IS a kinematics problem. Any equation used to solve it involves kinematics. An exact solution would require solving a (kinematic) differential equation.
A good approximation to part (a) can be obtained by assuming an average air resistance force of |V|/60 , where V is the initial velocity of 20 m/s. Thus work done against friction going up is
V/60*H, and initial kientic energy equals potential energy gain at the highest elevation PLUS work done against friction.
(1/2) M V^2 = M g H + V/60*H
H = (1/2)V^2/[g + V/(60M)]= (1/2)V^2/11
= 18.2 m
The time spent by the ball going up is
very nearly H/(V/2) = 0.46 s
Use a similar energy conservation argument to estimate the (longer) time it takes for the ball to come back down. Note that it must travel an additional 30 m to reach the ground.
how would you solve it with differential equations
Solve
M dv/dt = -M g -v/60*H
(v>0)
followed by
M dv/dt = -M g + v/60*H
(v<0)
Once you have v(t), integrate v dt to get the distance travelled vs t.
To find the maximum height the ball reaches and the time it takes to hit the ground, we can use the principles of energy conservation and consider the forces acting on the ball.
a. To find the maximum height above the ground the ball reaches, we first need to determine the initial potential energy of the ball when it is thrown from the roof.
The potential energy (PE) of an object at height h is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height above some reference point (in this case, the ground).
Given:
Mass of the ball (m) = 0.15 kg
Height above the ground (h) = 30 m
PE = mgh
PE = (0.15 kg) * (9.8 m/s²) * (30 m)
PE = 44.1 Joules
The initial potential energy of the ball is 44.1 Joules.
Now, let's consider the forces acting on the ball during its motion. There are two significant forces: the gravitational force and the force due to air resistance.
The gravitational force is always acting downwards and can be calculated using the formula Fg = mg, where Fg is the gravitational force and m is the mass of the object. In this case, Fg = (0.15 kg) * (9.8 m/s²) = 1.47 N.
The force due to air resistance is given by |v|/30, where v is the velocity of the ball.
When the ball is at its peak, the velocity is momentarily zero. At this point, the force due to air resistance is also zero.
Since the gravitational force is the only force acting on the ball when it reaches its maximum height, the gravitational potential energy is equal to the maximum kinetic energy of the ball.
Therefore, 44.1 Joules of potential energy is equal to the maximum kinetic energy when the ball reaches its maximum height.
The maximum kinetic energy (KEmax) of the ball can be calculated using the formula KE = (1/2)mv², where KE is the kinetic energy and v is the velocity of the ball.
KEmax = 44.1 Joules
Now we can solve for v using the equation 44.1 = (1/2)(0.15 kg)v²
v² = (2 * 44.1 Joules) / (0.15 kg)
v² = 588 Joules/kg
Taking the square root of both sides:
v = sqrt(588 Joules/kg)
v ≈ 24.25 m/s
Therefore, the maximum height above the ground the ball reaches is when its velocity is approximately 24.25 m/s.
b. To find the time it takes for the ball to hit the ground, we can use the principles of kinematics. However, since we are not allowed to use the kinematic equations, we will approach it differently.
When the ball hits the ground, its final height will be zero. We can use this information to determine the time it takes for the ball to hit the ground.
Using the equation of motion h = ut + (1/2)gt², where h is the height above the ground, u is the initial velocity, t is the time taken, and g is the acceleration due to gravity, we can set h = 0, take u as the negative of the initial velocity (since the ball is thrown upward), and solve for t.
0 = -(20 m/s)t + (1/2)(9.8 m/s²)t²
Rearranging the equation and setting it equal to zero, we have:
(1/2)(9.8 m/s²)t² - (20 m/s)t = 0
Simplifying,
4.9t² - 20t = 0
Dividing through by t, if t ≠ 0,
4.9t - 20 = 0
Solving for t,
t = 20 / 4.9
t ≈ 4.08 seconds
Therefore, the time it takes for the ball to hit the ground is approximately 4.08 seconds.