Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Anderson Video - Time for Dropped Object to Hit Ground

Professor Anderson
Was this helpful?
 >> Hello class, Professor Anderson here. Let's talk about dropping an object from the top of a building. Let's just drop it straight down. And let's say that we are standing up here at height H above the ground. And let's ask the following question: How long does it take to hit the ground? Now, we just looked at a problem where we tossed it off of the roof horizontally but now we want to convince ourselves that the time that we got there is the same as if we just dropped it straight down. So we go back to our kinematic equations and now we And now we can probably just plug in these numbers. Whenever you have a picture like this, you always have to identify some coordinate system. We'll say that y equals zero starts at the ground and goes up and so y equals H is in fact where we started. We're going to end up on the ground, we're going to start at height h, we're dropping it from rest. So that is zero. a y is of course negative g. And we get t equals the square root of two h over g. So exactly the same as we had before where we tossed the rock off horizontally. So this is the interesting thing about two dimension— two dimensional motion is that the motion is in fact decoupled. You have X equations that give you motion in the X direction and you have Y equations that give you motion in the Y direction. And for our projectiles we in fact knew the following. A sub x was equal to zero. A sub y was gravity, Negative G. So all those equations that we just learned in one dimension just apply again to two dimensions. Okay so let's go back to our original problem where we toss the rock off horizontally and let's figure out how far it goes.