es8515

A 3.49 kg ball is dropped from the roof of
a building 187.2 m high. While the ball is
falling to Earth, a horizontal wind exerts a
constant force of 13.2 N on the ball.
How long does it take to hit the ground?
The acceleration of gravity is 9.81 m/s
How far from the building does the ball hit
the ground?
.What is its speed when it hits the ground?

Woody

S=ut+½at²

This is an equation that crops up a lot in homework.

S="displacement" (the difference in the position of the object at time t relative to the position when t=0)
u ="initial velocity" (the velocity at time=0)
t = "elapsed time" (since S=0)
a = "acceleration" (for your question this is the acceleration due to gravity).

If you are happy with mathematical differentiation and integration,
then it is instructive to look at how integrating acceleration with respect to time gives velocity
(v=u+at) {assuming constant acceleration}
and integration of velocity (with respect to time) gives displacement
S=ut+1/2at^2

Note that ^2 is a way of expressing squared when we don't have the superscript 2 character available.

es8515

Ok so the time to hit the ground is 6,1809 s and it hits the ground 72.248 meters from the building.Still not sure how to figure out with what speed. I really truly appreciate your help.

benit13

Ok so the time to hit the ground is 6,1809 s and it hits the ground 72.248 meters from the building.Still not sure how to figure out with what speed. I really truly appreciate your help.
If you can, try and remember from memory the following equations. They are often known as the "SUVAT" equations. They basically describe the motion of an object under a constant acceleration:

$$\displaystyle v = u + at$$
$$\displaystyle v^2 = u^2 + 2as$$
$$\displaystyle s = ut + \frac{1}{2} at^2$$
$$\displaystyle s = \frac{1}{2}(u + v)t$$

In these equations:

s = displacement from origin (in m)
u = initial speed (in m/s)
v = final speed (in m/s)
t = duration (for object initially at speed u to reach speed v; in s)
a = acceleration (constant; in m/s2)

These are extremely useful equations since most terrestrial motion problems can be approximated as constant acceleration problems (e.g. by neglecting air resistance).

So... for question 1 you used $$\displaystyle s = ut+\frac{1}{2} at^2$$. For question 2, which formula will help you get the speed when the object hits the ground?

1 person

es8515

v=u+at ?
I am so stuck on it.
Sorry, I really appreciate it.

benit13

v=u+at ?
I am so stuck on it.
Sorry, I really appreciate it.
No worries!

Yes, try that formula.

Do you know u, a and t? If so, write them down:

u =... m/s
a = ... m/s2
t = ... s
v = ?

Then try substituting your numbers into the formula and see what you get.

es8515

v=187.2+(9.81x6.1809)
v=247.834 m/s

That just doesn't seem right.

benit13

You set your initial speed of the object to be 187.2 m/s... but that's the height of the building (187.2 m).

If you drop something, the initial speed is zero (it's only bigger than zero if you throw it!)

es8515

v=0+(9.81x6.1809)
v=60.634m/s

Is that correct?

Woody

Double Entendre

Sorry, I was in a bit of a rush for my initial answer,
I was just coming back to check how you were getting on,
thankfully Benit has filled in my gaps

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