Question

A stuntwoman 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?

Answer 1

Answer:

  She will make the jump.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$, 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

   $77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds$

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 $m/s^2$, time = 2.85

    Substituting

           $s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m$

  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.