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Question on Mechanics and projectile motion?
2 particles move in a uniform gravitational field with accelaration g. At the initial moment the particles were located at 1 point and moved with the velocities 3 m/s & 4 m/s resp. horizontally in opposite directions. Find the distance between them at the moment their velocity vectors become mutually perpendicular.
Call the projectile with 3m/s A and the other projectile B. Assume A moves to the left.
Imaging the two particles in a coordinate system. Assume that both A and B start at the origin. We can write equations relating their positions with time. Let A_x represent the x coordinate of A, and A_y represent its y coordinate, and the same goes for B. Let t = time measured in seconds.
A_x = -3t
A_y = -0.5gt^2 = -4.9t^2
B_x = 4t
A_y = -4.9t^2
In order to find when the two velocities are perpendicular, we look toward the slopes of these two graphs, which represent the velocity of the particle at a given time.
For A:
The velocity consists of a constant x component of -3m/s, and a changing y component of -9.8t. We can represent the direction of the velocity using a slope of these two components.
Slope = rise/run = 9.8t/3 = 3.26t
For B:
Similarly, the x component of velocity is always 4m/s, with the y component of velocity -9.8t again.
Slope = rise/run = 9.8t/-4 = -2.45t
To find when the two velocities are perpendicular, we find when the slopes are negative reciprocals of each other. So:
3.26t = -1/(-2.45t)
6.003t^2 = 1
t^2 = 0.167
t = 0.408
Now all that remains is to find the distance between the two points at time t = 0.408. Because the downward acceleration is constant, the two points are at the same height. So we find their horizontal distance which is
4t-(-3t) = 7t = 7*0.408 = 2.857 m
2.857 is the final answer
Uranya (2006)