A sandbag is dropped from the helicopter 125 m above the ground. What is its velocity at a height 60.0 m and when it hits the ground?
To find the velocity of the sandbag at a height of 60.0 m and when it hits the ground, we can use the equations of motion and the principle of conservation of energy.
Let's break down the problem into two parts - first, finding the velocity at a height of 60.0 m, and second, finding the velocity when it hits the ground.
1. Finding the velocity at a height of 60.0 m:
To find the velocity at this height, we can make use of the conservation of energy principle. The initial energy of the sandbag at a height of 125 m is given by its potential energy (PE) and the final energy at a height of 60.0 m is given by its kinetic energy (KE). The principle of conservation of energy states that the total energy of a system remains constant.
Initially, the potential energy is:
PE_initial = m * g * h_initial
where m is the mass of the sandbag, g is the acceleration due to gravity (9.8 m/s^2), and h_initial is the initial height (125 m).
Finally, the kinetic energy is:
KE_final = (1/2) * m * v_final^2
where v_final is the final velocity at a height of 60.0 m.
By equating the initial potential energy with the final kinetic energy, we can find the velocity at a height of 60.0 m.
PE_initial = KE_final
m * g * h_initial = (1/2) * m * v_final^2
Simplifying the equation, we get:
v_final = sqrt(2 * g * (h_initial - h_final))
Substituting the given values:
h_initial = 125 m
h_final = 60.0 m
g = 9.8 m/s^2
v_final = sqrt(2 * 9.8 * (125 - 60.0))
Calculate the final velocity using this equation.
2. Finding the velocity when it hits the ground:
To find the velocity when the sandbag hits the ground, we can use the kinematic equation:
v_final^2 = v_initial^2 + 2 * a * d
where v_initial is the velocity at a height of 60.0 m (which we have just calculated), a is the acceleration due to gravity (-9.8 m/s^2), and d is the distance or height fallen from the initial height (125 m).
Plugging in the values:
v_final^2 = v_initial^2 + 2 * (-9.8) * 125
Solve for v_final to get the velocity when it hits the ground.
Use conservation of energy
Penergy at start = m g h = m*9.81 * 125
Penergy at 60 = m g h = m*9.81*60
so
(1/2) m v^2 = m(9.81)(125-60)
then at the ground
Pe at ground = 0
(1/2) m v^2 = m (9.81)(125)