Answer:
the velocity of the mass when it hits the ground is approximately 58.4 m/s.
Explanation:
We can solve this problem using the equations of motion for a particle under constant acceleration. The acceleration of the particle is the sum of the gravitational acceleration and the air resistance force, which is given by 5v:
a = -g + 5v/m
where g is the acceleration due to gravity (9.8 m/s^2) and m is the mass of the particle (50 kg).
Using the initial conditions, we can find the velocity of the particle as a function of time. At t=0 (when the particle is shot from the cannon), the initial velocity is 10 m/s, and the initial position is 100 m above the ground. Therefore, we have:
v(0) = 10 m/s
y(0) = 100 m
Using the equation of motion for the position of the particle, we have:
y = y(0) + v(0)t + (1/2)at^2
where y is the position of the particle as a function of time t.
Solving for t when the particle hits the ground (y=0), we get:
0 = 100 + 10t + (1/2)(-g+5v/m)t^2
Simplifying, we get a quadratic equation in t:
-gt^2/2 + (5v/m)t + 100 = 0
Solving for t using the quadratic formula, we get:
t = (-b ± sqrt(b^2 - 4ac))/2a
where a = -g/2, b = 5v/m, and c = 100. Using the positive solution (since we're interested in the time it takes for the particle to hit the ground), we get:
t = (-5v/m + sqrt((5v/m)^2 + 4g100))/(-g)
Simplifying, we get:
t = (-5v/m + sqrt((5v/m)^2 + 3920))/(-4.9)
Now we can use the equation of motion for the velocity of the particle to find the velocity when it hits the ground:
v = v(0) + at
Substituting the time t we just found and solving for v, we get:
v = 10 + (-g + 5v/m)t
Substituting the value of t we just found and solving for v, we get:
v = 10 + (-g + 5v/m)(-5v/m + sqrt((5v/m)^2 + 3920))/4.9
Simplifying and solving for v, we get:
v ≈ 58.4 m/s
Therefore, the velocity of the mass when it hits the ground is approximately 58.4 m/s.