You throw a ball horizontally off of a building with a horizontal velocity of 10 m/s. What is the ball’s horizontal velocity 5 seconds later? What is the ball’s vertical velocity 5 seconds later?
This is a motion in two dimensions, which can be decomposed into two movements, which are the following:
. A uniform rectilinear motion in the horizontal direction(x), having a constant speed.
. A uniformly accelerated motion in a vertical direction(y), by the action of gravity
The initial velocity can be decomposed into a horizontal component vx = 10 m/s and a vertical component vy = 0.
Since, in the horizontal direction, the movement is at a constant speed, we have that, after 5 s, the horizontal speed is:
vx = 10 m/s
In the vertical direction we apply the following equation:
vx = v0x + gt
g, is the acceleration of gravity.
vx, is the velocity at any time.
v0x, is the initial velocity.
t, is the time.
Then we have that in the vertical direction, after 5s, speed is:
vx = 0 + (9.8)5 = 49 m/s
The horizontal velocity is constant for the entire motion until it hits the ground. Therefore, after 5 seconds, the ball will still have a velocity of 10m/s.
Next, we need to find the vertical component of velocity. First, assume a positive direction. Since we are throwing it from a building, we can choose down to be positive, meaning that up would be negative.
Now, use the formula `v=at ` and plug in what you know. A= 9.8m/s^2 (remember it is positive because we chose it to be), and t=5 seconds.
Plugging it all in, you get `v=(9.8m/s^2)(5s) `
Note that this assumes that the ball does not hit the ground AT 5 seconds.