Question 196885: We know how to find the time a falling object reaches the ground. We can use a similar technique to find the distance an object will travel, if it is launched, like a cannonball by a cannon (we assume no air friction). Let our cannon sit at x=0. Suppose it shoots a ball with an initial vertical velocity of v1, and horizontal velocity of v2. Physics will tell us that its trajectory will be given by the following argument: Vertical motion: y(t) = v1t -1/2gt^2 (we again assume y=0 is the ground). Horizontal motion: x(t) = v2t. Now we can eliminate t rrom these two equations, obtaining a relation between y and x (just solve for t in the second, and substitute t in the first with your expression from the second). The resulting function, y(x), tells us at what height the cannonball is, when it has traveled a distance x. Hence, the ball is on the ground when y(x) = 0. The equation will have two solutions in x: One will correspond to the starting position, when the ball was initially, and the other to the position at which it will land again. Solve the equation for the following values: g = 9.8 m/s^2. v1 = 10 m/s. v2 = 5 m/s.
Answer by solver91311(24713)   (Show Source):
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