Projectile motion example (time to impact, impact velocity, horizontal travel)

Projectile motion problem (ignoring air resistance). The time to impact, impact velocity, and horizontal travel are calculated. The interpretation of a “negative time” result (originally discarded in the “time to impact” calculation) is discussed.

[@miii336]

thank you thank you thank you

[@SN1350]

Thanks for the video, but maybe you could help me understand something that intuitively is far different than using formulas. Question 1. So, if you are standing at top of a building, let’s assume 200 ft up and you throw an object such as a rock in forward direction, let’s say 3 m/s, the rock will travel in a forward motion at first because it had a force pushing it forward and eventually it will no longer move forward and end up dropping to the ground in a straight line, especially if you are that high up in the air. We know that if it is falling straight down the velocity in the x direction is 0 so right before it hits the ground the velocity in x direction is 0. But at first it had to have some amount of velocity in the x direction created by the force when we threw the rock or it would not have moved forward, so why is the velocity in X direction in the beginning same as the velocity in X direction at the end? I know the formula says so, but in reality it makes no sense. Question 2. This one really beyond comprehension. I have watched some of these videos about projectile motion and people say that if you have two objects, one dropped straight down from 100 ft up and the other is thrown at 100 ft up, the will both hit the ground at the same time. Now how can that be? If an object is thrown it has to take some time to travel in the x direction before it hits the ground vs the other way where it wastes no time going in the X direction and hits the ground.

[@CraigWLooney]

QUESTION 1: the "projectile" analysis is only *after* the rock has left your hand. That is, the rock's initial 3 m/s horizontal velocity (in you question) is just *after* leaving your hand. So you are no longer exerting a horizontal force (so no horizontal acceleration so no change in horizontal velocity component). The force applied by your hand was needed to get the rock moving horizontally at 3 m/s in the first place, but once the rock becomes a "projectile" it has left your hand and so your hand no longer applies a force. QUESTION 2: An ideal projectile with an initial velocity that is completely horizontal does the following two things *at the same time*: [1] it continues to have a horizontal velocity component and that horizontal velocity component remains constant, and [2] its downward velocity component increases by an amount of 9.8 m/s every second. That is: the downward (component of the) motion begins *immediately* ... it does not wait to kick in. (From a force perspective: As soon as the rock leaves your hand, gravity is the only force acting on it ... so as soon as the rock leaves your hand, gravity accelerates it downward. Gravity doesn't care that the rock was initially moving horizontally ... gravity doesn't affect the horizontal motion because -- being a "downward" force -- it can only contribute a downward acceleration).

[@SN1350]

@@CraigWLooney Thanks for the reply. Amazing how one can learn a lot more from KZbin than sitting in a classroom.