Answer:
Explanation:
Use the one-dimensional equation
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] 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),
[tex]v_0=5[/tex], and
a = -9.8
Filling in:
[tex]-2=5t+\frac{1}{2}(-9.8) t^2[/tex] and simplified a bit:
[tex]-2=5t-4.9 t^2[/tex]
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:
[tex]-4.9 t^2+5t+2=0[/tex] (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.
