1. The problem statement, all variables and given/known data A small point-like object is thrown horizontally off of a 50 meter high building with an initial speed of 10 m/s. At any point along the trajectory there is an acceleration component tangential to the trajectory and an acceleration component perpendicular to the trajectory. How many seconds after the object is thrown is the tangential component of the acceleration of the object equal to twice the perpendicular component of the acceleration of the object? Ignore air resistance. 2. Relevant equations Kinematics? I can't seem to be able to mathematically express an "acceleration component perpendicular to the trajectory" 3. The attempt at a solution I know that the only force acting on the object is gravity so the acceleration vector would be downwards. However they're talking components, so I presume that this tangential component has an x-component that cancels out with the x-component of the perpendicular component. That way they yield a downward pointing vector...This problem shouldn't be too difficult, but I just don't understand how to approach it...btw the answer is 2.00 seconds.... I tried another approach (probably somewhat flawed) , which didn't delve into the mathematical aspect of the problem, but rather I just drew a diagram of the object's trajectory.... initially the acceleration is perpendicular to the starting point of the trajectory and when the object strikes the ground i would think the acceleration would be nearly tangential to the trajectory. however, this makes it sound like the trajectory is a circle (like the quarter of the circle that lies, in say the first quadrant)...so as you can see the overall acceleration vector, i hypothesized, would be more--- say 2 parts tangential to one part perpendicular...wow its hard to convey what i want to say...anyways, i feel like i am overthinking this problem and there should be some formulated, mechanical approach to get the answer. Any input/comment(s) are appreciated. Thank you for your help.