Aball is thrown straight up in the air. when will its kinetic energy be the smallest before it is caught? at the start of its flight halfway to the top of its flight at the top of its flight at the end of its flight

At the top of its flight.

At the top of the flight,the height from the ground is maximum. Owing to this position the potential energy attained here is the maximum.

Also potential energy = mgh

Where m is the mass of the object.

g is the acceleration due to gravity

h is the height of object.

From this equation we can conclude that at the top,the maximum height would be attained and hence it would possess maximum potential energy.

Also at the top the object will possess zero velocity.

Kinetic energy = 1/2 (mv^2).

Where m is the mass of the object

v is the velocity of the object.

Hence at the top, since the velocity of the object is zero,the kinetic energy would be zero as explained in the below equation.

Kinetic energy= 1/2(m x 0)

Kinetic energy = 0.

Thus at the top of the flight,the ball possess only potential energy and minimum (0) kinetic energy.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards