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Magnitude and Greatest Coordinates Acceleration

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A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 360 m at 40 degrees above the horizon. The airplane is tracked for another 123 degrees in the vertical east-west plane. The range at final contact being 790 m. Find the displacement of the airplane during the period of observation.

A particle leaves the origin with an initail velocity ( v - vector ) ( 3.00 I vector ) m/s and a constant acceleration ( a vector ) = ( -1.00 I vector - 0.500 j vector ) m/s^2 . When the particle reaches its maxium x coordinate, what are ( a ) its velocity and ( b ) its position vector ?

Particle A moves along the line y = 30 m with a constant velocity ( v vector ) of magnitude 3.0 m/s and directed parallel to the positive x axis. Particle B starts at the orgin with zero speed and constant acceleration vector a ( of magnitude 0.40 m/s^2 ) at the same instant that particle A passes the y axis. What angle between vector a and the positive y axis would result in a collision between these two particles ? ( If your computation involves an equation with a term such as t^4 , substitute u = t ^2 and then consider solving the resulting quadratic equation to get u )

A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown , where t = 0 at the instant the ball is struck. ( a ) How far does the golf ball travel horizontally before returning to ground level ? (B) What is the maximum height above the ground level attained by the ball ?

You throw a ball toward a wall with a speed of 25.0 m/s and at an angle of 40.0 degrees above the horizontal. The wall is 22.0 m from the release point of the ball. ( a ) How far above the release point does the ball hit the wall ? (b) What are the horizontal and vertical components of its velocity as it hits, has it passed the highest point on its trajectory ?

In Galileo's Two New Sciences the author states that evations ( angles of projection ) which exceed or fall short are in equal amounts the ranges are equal ....... Prove this statement

A baseball is hit at ground level. The ball reaches its maximum height above the ground level 3.0's after being hit. Then 2.5's after reaching its maximum height, the ball barely clears a fence that is 97.5 m from where it was hit. Assume the ground level is reached by the ball ? ( b ) how high is the fence? ( c ) How far beyond the fence does the ball strike the ground ?

You are to ride jet cycle over a lake, starting from point I. First moving at 30 degree's north of due east :

1. Increase your speed at 0.400 m/s^2 for 6.00 s
2. With whatever speed you then have, move for 8.00's
3. Then slow at 0.400 m/s^2 for 6.00 s

Immediately next , moving due west:

4. Increase your speed at 0.400 m/s ^2 for 5.00 s.
5. With whatever speed you then have, move for 10.0 s.
6. Then slow at 0.400 m/s^2 until you stop.

In magnitude - angle notation what then is your average velocity for the trip from point I ?

A ball is launched with a velocity of 10 m/s, at an angle of 50 degree's of the horizontal. The launch point is at the base of a ramp of horizontal length d1 = 6.00 m and height d2 = 3.00 m. A plateau is located at the top of the ramp. (a) does the ball land on the ramp or the plateau ? When it lands, what are ( b ) the magnitude and ( c) the angle of its displacement from the launch point ?

A cart is propelled over an xy plane with acceleration components a x = 4.0 m/s^2 and a y = - 2.0 m/s^2. Its initial velocity components v 0 x = 8 m/s and v 0 y = 12 m/s. In unit vector notation what is the velocity of the cart when it reaches its greatest coordinate ?

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Solution Summary

The magnitude and greatest coordinate acceleration is examined. The components are provided for the unit vectors. Horizontal degrees are discussed.

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Problem 6 : because range = u u*sin(2x)/g
It will maximum when 2x = 90 or x =45 degree
Also sin 2x = sin (180 - 2x)

Therefore range for the angles x and 90 - x will be equal

Problem 5:
horizontal vel. = 25 cos(40) = 19.15 m/s
Vertical vel. = 25 sin 40 = 16.07 m/s
Time to hit the ball t = 22/19.15 = 1.15 sec

Vertical distance traveled = 16.07*1.15 - ½* 9.81*1.15*1.15
= 18.48 - 6.40
= 12.08 m
at highest point vertical component will be zero.
And horizontal component will be same ...

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