The mass is moving by uniformly accelerated motion, with initial velocity

and acceleration

. Its position at time t is given by the following law:

where we take the initial position

since we are only interested in the distance traveled by the mass.
If we put

into the equation, the corresponding time t is the time it takes for the mass to travel this distance:


And the two solutions for the equation are:

--> negative, we can discard it

--> this is the solution to our problem
Answer:
0.32 m.
Explanation:
To solve this problem, we must recognise that:
1. At the maximum height, the velocity of the ball is zero.
2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.
With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:
Mass (m) = constant
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) = 2.5 m/s
Height (h) =?
PE = KE
Recall:
PE = mgh
KE = ½mv²
Thus,
PE = KE
mgh = ½mv²
Cancel m from both side
gh = ½v²
9.8 × h = ½ × 2.5²
9.8 × h = ½ × 6.25
9.8 × h = 3.125
Divide both side by 9.8
h = 3.125 / 9.8
h = 0.32 m
Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.
Explanation:
It is given that,
Mas of the object, m = 6 kg
It is lifted through a distance, h = 5.25 m
Tension in the string, T = 80 N
(a) By considering the free body diagram of the object, the forces can be equated as :




Work done by tension, 


(b) Work done by gravity, 


(c) Let v is the final speed of the object and u = 0


v = 5.91 m/s
Hence, this is the required solution.
Explanation:
Given
mass of the rock is 10 kg
Force requires to hold the rock is equal to its weight
Weight is given by the product of mass and acceleration due to gravity
Weight on the earth surface

Weight on the moon surface

So, the force holding the rock on earth is approximately 6 times the force on the moon.