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.

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

talk back to him about the sport and show him your expertise or leave the room