# Answer to Question #1529 in Mechanics | Relativity for noor

Question #1529

A 0.20 kg stone is held 2.0 m above the ground and then dropped .
1- find the stone's total mechanical energy at the point where it is dropped.
2- find the stone's total mechanical energy when it reached to the ground.
3- find the stone's speed at a height of 0.7m from the ground .
4- find the stone's speed when it reached to the ground .
5- find the height at which the stone's speed is 3.0 m/s

Expert's answer

1. The total stone's mechanical energy is equal to its initial potential energy: mgh = 0.2 * 9.8 * 2.0 =

2. Due to the law of energy concervation the total mechanical energy remains constant during the motion of the stone. E =

3. According to the equations of the motion,

h =h

2-0.7 = 9.8t2/2

t= 0.51 s.

V = 0.51*9.8 =

4. h =h

2 = 9.8t2/2

t = 0.64 s

V = 0.64*9.8 =

5. 3 = 9.8t, t = 0.31s.

h =h

**3.92 J**.2. Due to the law of energy concervation the total mechanical energy remains constant during the motion of the stone. E =

**3.92 J**.3. According to the equations of the motion,

h =h

_{0}+ v_{0}t + at^{2}/2, v=v_{0}+ at the positive axis is directed upward, a=g is negative, v_{0}= 0, h_{0}=2, h = 0.7. Let's find t.2-0.7 = 9.8t2/2

t= 0.51 s.

V = 0.51*9.8 =

**5.05 m/s**.4. h =h

_{0}- gt^{2}/2, h=0:2 = 9.8t2/2

t = 0.64 s

V = 0.64*9.8 =

**6.26 m/s**5. 3 = 9.8t, t = 0.31s.

h =h

_{0}- gt^{2}/2 = 2 - 9.8*0.312/2 =**1.54 m**.
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