(a) The maximum height on Dweeb is one third that on Earth.
(b) The maximum height is three times that on Earth.
(a) When the dissipative forces are absent, the total mechanical energy of a system remains constant.
If the ball of mass m is given a velocity v, then its kinetic energy is given by,
[tex]E_K=\frac{1}{2} mv^2[/tex]
As the ball moves upward, its velocity decreases due to the action of the gravitational force acting on the ball. When its velocity becomes zero, the ball would have reached its maximum height. Therefore, at maximum height, its entire kinetic energy transforms into potential energy.
On Earth, if the ball reaches a maximum height h and the acceleration due to gravity on Earth is g, then,
[tex]\frac{1}{2} mv^2=mgh[/tex]......(1)
If the ball is given the same velocity on Dweeb, it has the same kinetic energy as the ball on Earth. The gravitational force g ₁on Dweeb is 3 times that on Earth. Therefore,the potential energy of the ball at its maximum height h₁ on Dweeb is given by,
[tex]mg_1h_1=m(3g)h_1=3mgh_1[/tex].......(2)
Since the ball has the same kinetic energy at the beginning of the throw,
[tex]\frac{1}{2} mv^2=3mgh_1[/tex]......(3)
Equate equations (1) and (3)
[tex]mgh=3mgh_1\\ h_1=\frac{h}{3}[/tex]
The ball rises to one- third of its height on Earth if it is thrown with the same velocity on Earth and on Dweeb.
(b) If the ball is thrown on Dweeb with 3 times its velocity on Earth, then its kinetic energy at the start of the throw is given by,
[tex]E_K=\frac{1}{2} m(3v)^2=9(\frac{1}{2}mv^2)......(4)[/tex]
If it rises to a maximum height h₂on Dweeb, then,
[tex]mg_1h_2=9(\frac{1}{2}mv^2)......(5)[/tex]
From equation (1)[tex]\frac{1}{2} mv^2=mgh[/tex], and using g₁=3g
[tex]mg_1h_2=9(\frac{1}{2}mv^2)\\ m(3g)h_2=9mgh\\ h_2=3h[/tex]
If the ball is projected on Dweeb with 3 times the velocity on Earth, it would rise to 3 times the maximum height it reaches on Earth.
