Physics Question?

Question

A Baseball is dropped from rest from the top of a roof and fell 10 meters to the ground. Assume up is the positive direction. What is the average velocity of the baseball during the free falling motion?

The correct answer is actually -7 m/s. It was a mid-term question that I missed. I keep getting -14 m/s...

Answer 1

Let'slearntothink has made a mistake in the following:
"= 1/2 a t^2; s = 10 m, a = 9.8 m/s^2; t^2 = 2/9.8 = 20/98 = 10/49
t = sqrt(10/49) = 0.451 s"
This is not the correct value for t, see below for the correct one.

Short answer:
-7.0m/s

How I got this answer:
This is assuming no air friction.

We need to use the right formula to find how long it took the baseball to fall. Use

Δd=viΔt + ½a(Δt)²

Δd is change in position, vi is initial velocity, a=acceleration, Δt is the change in time(all are in SI units)

Plug in what you know and solve.

Δd=viΔt + ½a(Δt)²
-10m=0+½(-9.8m/s²)(Δt)²
-10m=-4.9m/s²(Δt)²
(Δt)²=2.041s²
Δt=-1.429s

Now look at the definition of average velocity.

Vavg=Δd/Δt

Vavg=avg velocity, Δd=total displacement, Δt=total change in time(all are in S.I. units)

Plug in what we know.
Vavg=Δd/Δt
Vavg=-10m/1.429s
Vavg=-7.0m/s

Answer 2

It's good to learn from your mistakes. So to understand what went wrong with your solution:

You calculated the velocity after a 10 m drop. Perhaps using the formula
V^2 = Vo^2 + 2*a*s where a = -g = -9.8 m/s^2
V will come out as -14 m/s. That's an instantaneous value for the velocity.

To average a collection of values you add them up and divide by the number of them. The increase in velocity here is linear - same acceleration during the drop. So take the value at the start, 0 m/s; and at the end, -14 m/s; add 'em up and divide by 2.

Answer 3

Neglecting air resistance, baseball falls with constant acceleration 9.8 m/s^2.

s = 1/2 a t^2; s = 10 m, a = 9.8 m/s^2; t^2 = 2/9.8 = 20/98 = 10/49

t = sqrt(10/49) = 0.451 s

average velocity = s/t = 10/0.451 = 22.17 m/s