Answer:
Explanation:
This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".
In the y-stuff category:
v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)
Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)
a = -9.8 m/s/s
t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)
In the "x-stuff" category:
v₀ = 7.25 m/s (this is given)
Δx = ???
a = 0 (acceleration in this dimension is ALWAYS 0)
t = (we will solve for this in the y-dimension and plug it in here).
In the y dimension:
Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and plugging in from the y-dimension info:
[tex]-10.0=0t+\frac{1}{2}(-9.8)t^2[/tex] which simplifies to
[tex]-10.0=-4.9t^2[/tex] so
[tex]t=\sqrt{\frac{-10.0}{-4.9} }[/tex] which, to 2 significant digits is
t = 1.4 seconds
Now we will do the same in the x-dimension, using t = 1.4:
Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and filling in the x-stuff:
Δx = [tex]7.25(1.4)+\frac{1}{2}(0)(1.4)^2[/tex] Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:
Δx = 7.25(1.4) + 0 so
Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).
