a squirt gun is tied to a string and released when it is 1 meter above the ground. the mass of the gun is .1kg. what ist its initial PE? 1Joule

what is its PE when it has fallen to .75meters? .75 J
What is its KE at this point? .25 J
What is its velocity? 0 m/s

Above is the correct answer I solved, but help me using the above information, to these questions.

what is its PE when it has ffallen to 0.5 m ?
what is its KE at his point?
What is its velocity?

what is its PE when it has fallen to 0.25m?
what is its KE at this point?
what is its velocity?
and what is its velocity right before it hits the ground?

To find the potential energy (PE) when the squirt gun has fallen to a certain height, you can use the formula: PE = mgh, where m is the mass of the gun (0.1 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height.

1. When the squirt gun has fallen to 0.5 meters:
PE = (0.1 kg)(9.8 m/s^2)(0.5 m) = 0.49 J (Joules)

To find the kinetic energy (KE) at the given point, you can use the formula: KE = 0.5mv^2, where m is the mass of the gun (0.1 kg), and v is the velocity.

2. KE at 0.5 meters:
KE = 0.5(0.1 kg)(0 m/s)^2 = 0 J

The velocity (v) at this point is zero because the squirt gun is not moving.

3. When the squirt gun has fallen to 0.25 meters:
PE = (0.1 kg)(9.8 m/s^2)(0.25 m) = 0.245 J

4. KE at 0.25 meters:
KE = 0.5(0.1 kg)(0 m/s)^2 = 0 J

Again, the velocity (v) at this point is zero.

Right before the squirt gun hits the ground, its potential energy will be zero since it has fallen completely, and all the energy has been converted into kinetic energy.

5. Velocity right before hitting the ground:
Since the potential energy is now zero and the kinetic energy is nonzero, we can equate the two:
0.5(0.1 kg)v^2 = 0.245 J (the value of PE at 0.25m)
v^2 = (2 * 0.245 J) / 0.1 kg
v^2 = 4.9 m^2/s^2
v ≈ 2.21 m/s

Therefore, the velocity right before the squirt gun hits the ground is approximately 2.21 m/s.