Atoy rocket is launched from a 2.8 m high platform in such a way that its height, h (in meters), after t seconds is given by the equation h equals negative 4.9 t squared plus 18.9 t plus 2.8. how long will it take for the rocket to hit the ground?

3.98 s (approximately 4 s for practical purposes)

Explanation:

1) Equation given:

h(t) = - 4.9 t² + 18.9 t + 2.8

2) The time it will take the rocket fo hit the ground is given doing h (t) = 0

- 4.9 t² + 18.9 t + 2.8 = 0

3) You can solve that equation using the quadratic equation or working the equation to complete squares.

To use the quadratic equation you have a = - 4.9, b = 18.8, and c = 2.8

With that, t = [-18.8 +/- √ ( 18.8² - 4(-4.9)(2.8) ) ] / (2(-4.9) ) = 3.98 s

(I only picked the positive result because the negative does not have physical meaning).

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