Astuntwoman is going to attempt a jump across a canyon that is 77 m wide. the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave. the takeoff ramp is built with a 15º angle from horizontal. if the stuntwoman leaves the ramp with a velocity of 28 m/s, will she make the jump? why?

  She will make the jump.

Explanation:

We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

   Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

So she will cover 77 m in 2.85 seconds

 Now considering vertical motion, up direction as positive

    Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 , time = 2.85

    Substituting

  So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

  So she will make the jump.

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.