A woman throws a cricket ball upwards into the air. The ball leaves the womans hand at 1.8m above the ground at speed of 9.7. It rises and then falls back to the ground. Considering the energies involved calculate the speed at the point it hits the ground.
accerlation due to gravity (g) is 9.8 m s, neglect air resistance
Equate final kinetic energy to initial kinetic and potential energies.
Initial:
KE=(1/2)m(9.7)²
PE=mg(1.8)
Final:
KE=(1/2)mv²
PE=0
All energies are in joules if you work with metres, kilograms and seconds.
Is that the same as
Ek = ½mv² - ½mu²
or does mgh have to be in the equation?
mgh has to be in the equation because the ball was initially 1.8m higher than the ground. The velocity required is at ground level.
(1/2)mv²=(1/2)mu²+mgh
u,v being initial and final velocities.
To calculate the speed at which the ball hits the ground, we need to consider the conservation of energy.
The initial kinetic energy of the ball as it leaves the woman's hand is given by 1/2 * m * v^2, where m is the mass of the ball and v is its initial velocity.
As the ball rises, it gains potential energy due to its increasing height. The potential energy at any height h is given by m * g * h, where g is the acceleration due to gravity.
At the peak of its trajectory, the ball's velocity is momentarily zero. At this point, all of its initial kinetic energy has been converted to potential energy. The maximum height h can be determined using the equation:
m * g * h = 1/2 * m * v^2
Simplifying, we find:
h = v^2 / (2 * g)
The total energy of the ball is constant, so at the point it hits the ground, all of its potential energy has been converted back to kinetic energy.
Setting the potential energy equal to kinetic energy, we have:
m * g * h = 1/2 * m * v_final^2
Substituting the value of h we found earlier:
v_final^2 = 2 * g * h
Solving for v_final, we have:
v_final = sqrt(2 * g * h)
Plugging in the values given in the question, where g = 9.8 m/s^2 and h = 1.8 m, we can calculate the final velocity:
v_final = sqrt(2 * 9.8 * 1.8) m/s
v_final ≈ 8.8 m/s
Therefore, the speed at which the ball hits the ground is approximately 8.8 m/s.