Since there's no air resistance, this is a problem in projectile motion. The goal is to find the point where the ball's parabolic trajectory intersects the incline. It is best to choose the xx-axis to be horizontal and direct to the right, the yy-axis to be vertical and direct to the up, and the origin to be at the point where the ball is fired. In the projectile equations, the launch angle α0α0 is measured from the horizontal. What is this angle in terms of θθ and ϕϕ?

good for her? ? where is the question?

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

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2