Question

6) An airplane is flying over the countryside with a velocity of 85.0 m/s at an angle of 19.5o with respect to the horizontal. The airplane releases a crate of supplies when it is 85 m directly above a person standing on the ground. How far from the person will the crate land?

7) A motorcycle stunt rider rides horizontally off the edge of a vertical cliff of height 8.5 m and lands safely 12.3 m from the base of the cliff.
a) With what velocity did the motorcycle leave the cliff?
b) What was the direction of the motorcycle

Answer 1

(6) The position of the crate from the person when released from moving ariplane can be determined as follows.

The crate is released from flying airplane with velocity of 85.0 m/s at an angle of 19.5o . Thus, the velocity of airplane in horizontal direction and in vertical direction are:

The crate released from airplane so it has same velocity. Thus, the time taken by crate to reach the ground is,

           

           

On solve for t; the roots are 7.967 and -2.177.

The time never be negative, thus t = 7.967 s.

Therefore, the horizontal displacement of the crate is,

    

(7) A motorcycle stunt rider rides horizontally off the edge of a vertical cliff of height 8.5 m and lands safely 12.3 m from the base of the cliff.

(a) The velocity of the motorcycle when leave the cliff can be determined as follows:

The height of the cliff is 8.5 m. The time taken to reach the ground from the cliff is,

         

since initially the rider have only horizontal velocity, thus the initial vertical velocity is zero.

Thus,

Rearrange the equation for t.

       

The initial horizontal velocity of the rider is,

          

(b) The direction of the motorcycle before hitting the ground can be determined as follows:

The final velocity of the rider is,

              

             

              

The velocity components when the rider hitting the ground are and .

Therefore, the direction of the velocity is,

                or 54.1 degress below the positive x-axis.

(8)

A motorcycle stunt rider rides horizontally off the edge of a vertical cliff of height H and lands safely D from the base of the cliff.

(a) The velocity of the motorcycle when leave the cliff can be determined as follows:

The height of the cliff is 8.5 m. The time taken to reach the ground from the cliff is,

since initially the rider have only horizontal velocity, thus the initial vertical velocity is zero.

Thus,

Rearrange the equation for t.

The initial horizontal velocity of the rider is,

(b) The direction of the motorcycle before hitting the ground can be determined as follows:

The final velocity of the rider is,

The velocity components when the rider hitting the ground are and .

Therefore, the direction of the velocity is,

(9) The minimum speed required for the ball to clear the 90.0 cm-high net 16.0 m from the server if the ball is “launched”from a height of 2.65 m can be determined as follows:

The distance of net top from the point of impact = 2.65 m -0.90 m = 1.75 m

The time taken by ball to reach the net is

       
The horizontal velocity is,

           

10) Assume the Earth’s orbit around the sun is a circle and has radius 1.50 x 108 km. This is a rough approximation of the Earth’s true elliptical orbit.

(a) The orbital velocity of the earth is,

       

(b) The radial acceleration of the earth is,