![]() ![]() Hard Solution Verified by Toppr An object is thrown with a velocity v o with an angle with horizontal x-axis. Let’s consider a projectile launched at an angle with an initial velocity v0. Derive the equations of projectile motion. The motion occurs in a uniform gravitational field. ![]() ![]() To derive the equations of projectile motion, we’ll assume the following: The only force acting on the projectile is gravity. Imagine if you had to rederive the Pythagorean theorem every time you wanted to use it instead of just being able to plug the numbers into the formula. It follows a curved path known as a parabola. Also, once you have a general expression for a thing, you've essentially solved that class of problem. In general, whenever you can – that is, whenever it's not prohibitively difficult – you should try to solve the thing symbolically to gain the greatest insight. There are three equations of motion that can be used to derive components such as displacement (s), velocity (initial and final), time (t) and acceleration (a). For example, Maybe the expression for the area of a circle shows up somewhere in the final expression, which can suggest a different derivation or interpretation. But when you solve the thing symbolically, you can interpret the equation, see clearly what's proportional to what, any algebraic symmetry (functional symmetry, being able to swap variables, so on), you can see patterns or that some other quantity might be hidden in the thing. When you solve a thing numerically, you just get some number (or a vector, etc.) at the end (and maybe some units). Yeah, and it's actually a great way to gain insight into the nature of the thing. + y B y A + v Ay t AB 0.5g c t AB 2 v Ay. + x B x A + v Ax t AB and v Ax 15 cos 40m/s Now write a vertical motion equation. Then write the equation for horizontal motion. 8 s m ) 2 (plug in horizontal and vertical components of the final velocity) v, squared, equals, left parenthesis, 7, point, 00, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, right parenthesis, squared, plus, left parenthesis, minus, 20, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, right parenthesis, squared, start text, left parenthesis, p, l, u, g, space, i, n, space, h, o, r, i, z, o, n, t, a, l, space, a, n, d, space, v, e, r, t, i, c, a, l, space, c, o, m, p, o, n, e, n, t, s, space, o, f, space, t, h, e, space, f, i, n, a, l, space, v, e, l, o, c, i, t, y, right parenthesis, end text Solving the two equations together (two unknowns) yields R 42.8 m t AB 3.72 s Solution: First, place the coordinate system at point A. V 2 = ( 7.00 m s ) 2 + ( − 20.8 m s ) 2 (plug in horizontal and vertical components of the final velocity) v^2=(7.00 \dfrac v 2 = ( 7. Derivation of equations projectile motion homework-and-exercises kinematics projectile 24,336 Solution 1 We always start from the four kinematic motion equations: s s 0 + v 0 t + 1 2 a t 2 s s 0 + 1 2 ( v 0 + v) t v v 0 + a t v 2 v 0 2 + 2 a ( s s 0) These equations count along any path. ![]()
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