A 0.15-kg ball dropped from a helicopter at a height of 1.0 x 102 m. Show that the speed of the ball is about 44 m/s just before it hits ground (g = 9.8 m/s2 and ignore air
resistance).
gravitational potential energy becomes kinetic energy
1/2 m v^2 = m g h ... v = √(2 g h) = √(2 * 9.8 * 100)
To calculate the speed of the ball just before it hits the ground, we can use the equation for gravitational potential energy.
Gravitational potential energy (PE) is given by the formula:
PE = mgh
Where:
m = mass of the ball (0.15 kg)
g = acceleration due to gravity (9.8 m/s2)
h = height from which the ball is dropped (100 m)
Substituting the given values into the formula, we have:
PE = (0.15 kg)(9.8 m/s2)(100 m)
PE = 147 J
The potential energy is equal to the kinetic energy just before hitting the ground.
Kinetic energy (KE) is given by the formula:
KE = 1/2 mv^2
Where:
m = mass of the ball (0.15 kg)
v = velocity/speed of the ball just before it hits the ground
Substituting the given value of PE into the formula, we have:
147 J = 1/2 (0.15 kg) v^2
Simplifying the equation:
294 J = 0.15 kg v^2
Dividing both sides by 0.15 kg:
v^2 = (294 J) / (0.15 kg)
v^2 ≈ 1960 m^2/s^2
Taking the square root of both sides to solve for v:
v ≈ √(1960 m^2/s^2)
v ≈ 44.27 m/s (rounded to two decimal places)
Therefore, the speed of the ball just before it hits the ground is approximately 44 m/s.
To determine the speed of the ball just before it hits the ground, we can use the principles of kinematics and the equation for free fall motion.
The equation we'll use is:
v^2 = u^2 + 2as
where:
v is the final velocity (speed of the ball just before it hits the ground),
u is the initial velocity (0 m/s since the ball is dropped),
a is the acceleration due to gravity (9.8 m/s^2), and
s is the distance covered (the height from which the ball is dropped).
Given:
u = 0 m/s
a = 9.8 m/s^2
s = 1.0 x 10^2 m
Plugging these values into the equation:
v^2 = 0^2 + 2 * 9.8 * 1.0 x 10^2
Simplifying:
v^2 = 0 + 19.6 x 1.0 x 10^2
v^2 = 1960
Taking the square root of both sides:
v ≈ √1960
v ≈ 44.27 m/s
Therefore, the speed of the ball just before it hits the ground is approximately 44 m/s (rounded to two decimal places).