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Two balls are thrown simultaneously from the top of a tower with the speed of 9 m/s and 4 m/s in mutually opposite directions. What is the distance between them when their velocities are perpendicular to each other?
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Two balls are thrown simultaneously from the top of a tower with the speed of 9 m/s and 4 m/s in mutually opposite directions. What is the distance between them when their velocities are perpendicular to each other?
Sorry, I don’t do someone else’s homework. I’ll give you some pointers on how I would solve this problem, though.
First, the deal about their velocities (= vector representation of their speeds, not just their magnitude) being perpendicular seem to imply that you would need to solve a system of trigonometric equations, one for each object, so their angles are perpendicular to each other.
But, there is another way. In Cartesian geometry, we-re taught that two vectors (a1, b1) and (a2, b2) are perpendicular to each other if the following is true:
a1/b1 = -b2/a2.
So, no trigonometric functions needed. All that is left is to replace a’s and b’s for objects 1 and 2 with their motion equations in the x and y coordinates, and solve for the resulting messy-looking equality (make sure you do mind your signs!). That, in turn, is just drudgery once you know the equations for 2-dimensional or projectile motion. Once again, mind the signs!
As usual when working with ratios, you need to watch out that none of the denominators become zero, or whatever answer you get will be invalid. To get out of that conundrum, you may simply swap denominators and numerators so the zeros go to the numerators instead. The identity would still be valid, though:
b1/a1 = -a2/b2