Answer:
1.628 [tex]\frac{m}{sec^{2} }[/tex]
Explanation:
Anywhere in the universe, In a closed system, Conservation of energy is applicable.
In this case
Neil is initially on the surface of moon and has a velocity of 1.51 [tex]\frac{m}{sec}[/tex] in upward direction.
⇒He has Kinetic energy= [tex]K_{i}[/tex] = [tex]\frac{1}{2} m{v^{2} }[/tex] J
But with respect to the surface of the moon,
where m=mass of moon
v=velocity of Neil
He has Potential energy= [tex]P_{i}[/tex]=0 J
At the highest point of his jump, his velocity =0
⇒ Kinetic energy=[tex]K_{f}[/tex]=0 J
His Potential energy with respect to the surface of moon=[tex]P_{f}[/tex]=[tex]m \times g\times h[/tex]
where m=mass of moon
g= gravitational acceleration on moon
h=height from moon's surface
By Conservation Energy Principle
[tex]K_{i}[/tex]+[tex]P_{i}[/tex]=[tex]K_{f}[/tex]+[tex]P_{f}[/tex]
[tex]K_{i}[/tex]+0=0+[tex]P_{f}[/tex]
[tex]\frac{1}{2} m{v^{2} }[/tex] = [tex]m \times g\times h[/tex]
[tex]\frac{v^{2} }{2}[/tex] = [tex] g\times h[/tex]
[tex]\frac{1.5^{2} }{2}[/tex] J= [tex] g\times 0.7 m[/tex]
⇒ g = [tex]\frac{1.14}{0.7}[/tex] = 1.628 [tex]\frac{m}{sec^{2} }[/tex]
