Question

a ball rolls off the edge of a 2m high shelf at a speed of 5 m/s and hits the ground the taken to hit the ground is

Answer 1

Answer:

Explanation:

Use the one-dimensional equation

Δx = $v_0t+\frac{1}{2}at^2$ where delta x is the displacement of the object, v0 is the velocity of the object, a is the pull of gravity, and t is the time in seconds. That's our unknown.

Δx = -2 (negative because where it ends up is lower than the point at which it started),

$v_0=5$, and

a = -9.8

Filling in:

$-2=5t+\frac{1}{2}(-9.8) t^2$ and simplified a bit:

$-2=5t-4.9 t^2$

this should look hauntingly familiar (a quadratic, which is parabolic motion...very important in physics!!). We begin by getting everything on one side of the equals sign and solving for t by factoring:

$-4.9 t^2+5t+2=0$ (the 0 is also indicative of the object landing on the ground! Isn't this a beautiful thing, how it all just works so perfectly together?)

When you factor this however your math/physics teacher has you factoring you will get that

t = 1.3 sec and t = -.31 sec

Since we all know that time can NEVER be negative, it takes the ball 1.3 sec to hit the ground from a height of 2 m if it is rolling off the shelf at 5 m/s.