It's the equation of motion I described above.ĭon't worry if you haven't learned about integration yet the only thing you need to worry about is the power of #t# as we move from acceleration to velocity to position. Of course, this equation will probably look familiar to you. The constant of integration here is interpreted to be initial velocity, so I've just named it #v_i# instead of #C#.Īgain, the constant of integration is interpreted in this case to be initial position. ( #a = k# is interpreted as being #a = kt^0#) We can move on to velocity by integrating with respect to #t#. However, I can explain a bit more in-depth why this works, if you'd like, by doing a little integration. Is quadratic, and therefore describes a parabola. Put simply, basic projectile motion is parabolic because its related equation of motion, is also assumed) and also because horizontal velocity is generally constant. Projectile motion is parabolic because the vertical position of the object is influenced only by a constant acceleration, (if constant drag etc. So where,the horizontal component of velocity pushes it forwards,vertical component of velocity pushes it upwards,but it comes back to the ground just because of the gravitational force. So, at the highest point of its motion,the projectile has no vertical component of velocity,only horizontal component of velocity exists.Īfter that the projectile starts coming down being accelerated by gravity.So it's height decreases and at a time it reaches the ground. So,vertical component of its velocity is #u sin theta#,so the projectile will keep on moving up,untill it's upward velocity becomes zero due to the downward direction of gravitational force acting on it. Now suppose a projectile is projected with an initial velocity #u# at an angle #theta# w.r.t the horizontal,so we can discuss the effect of gravity after breaking the velocity into two perpendicular components.Īs gravity will affect the vertical component only. 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.Gravity opposes the vertical component of velocity of the projectile with which it has been projected. 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. 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. 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
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