3D Kinematics
In the -coordinate system, a massive particle is initially at rest on a plane whose equation is given below: The particle is launched with initial velocity . It travels under the influence of a gravitational acceleration of until it intersects the plane once more.
When the particle intersects the plane for the second time, how far away is it from the launch point (to 3 decimal places)?
The answer is 26.484.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Let the particle be thrown from the origin. Since there is no acceleration in the x and the y directions, therefore, the x and y coordinate of the particle after time t will be, x = − 3 t , y = 7 t . There is an acceleration of -10 units in the z direction. Therefore, using the equation of motion we may write the z coordinate as z = 1 2 t − 2 1 1 0 t 2 .
When the particle lands back on the plane, these coordinates will satisfy the equation of the plane. Hence, − 3 t + 1 4 t + 3 6 t − 1 5 t 2 = 0 ⇒ t = 1 5 4 7 .
After time t , the distance of the particle from the origin (the launch point) is d = x 2 + y 2 + z 2 = ( − 3 t ) 2 + ( 7 t ) 2 + ( 1 2 t − 2 1 ⋅ 1 0 t 2 ) 2 .
Substituting the value of t , we will get d = 2 6 . 4 8 4 .