How Fast is a Golf Ball Traveling 10m After Being Dropped from a Cliff?

[Michael17]

Golf balls and cliffs!

Hi I'm new to this forum and a novice in physics could anyone please help me with this question?

A golf ball is droped from the top of a sheer cliff 78m high. If gravity is 9.8ms^-2 and there is no air resistance, what speed is it traveling at 10m from its initial position?

 

[timmay]

Please try using the following formulae:

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

You have the initial speed as it's dropped from rest; you have a value for the acceleration due to gravity; you have the displacement that it travels. One of the 1D kinematics equations will be able to help you.

 

[Georgepowell]

What you should do is write all the information that you have, or you want, in a list.

u = 0 (initial velocity)
v = v (final velocity, we don't know this)
a = 9.8 (acceleration)
s = 10 (distance traveled)
t = ? (time taken, we don't need to know this)

then use the formula:

v²=u²+2as (then put in that values you know)
v²=0+2*9.8*10
v²=196
v= 14 m/s

The other formulas that you might need are:

v = u + at
s = 0.5t(u+v)
s = ut +0.5at²
v² = u² + 2as

 

[m.e.t.a.]

Hi, welcome to PF. Just to let you know, there is a homework help section for these kinds of questions. If you have more homework questions in future, you'll get the most helpful support if you ask them there.

Ok, so for this question you could just insert your values of [itex]a[/itex] and [itex]s[/itex] into whichever formula seems to fit, as is too often taught in schools, and in doing so stumble across the correct answer. But you learn much more if you ignore the formulas and work out the answer step by step.

Here, [itex]u[/itex] = 0 m/s, [itex]a[/itex] = 9.8 m/s[itex]^2[/itex], [itex]s[/itex] = 10 m and [itex]v[/itex] is the unknown. (At this point you could simply use [itex]v^2=u^2+2as[/itex], but please abstain from doing that until you're satisfied you understand where that formula from.)

The most intuitive way to visualise these types of question (in my opinion) is to introduce time. It's just easier to visualise what will happen to the ball/projectile over time, rather than over distance.

I'd start by expressing [itex]v[/itex] in terms of [itex]a[/itex] and [itex]t[/itex] ([itex]u[/itex] is zero so you can ignore it. Zero plus anything is zero.) Next: given that your formula contains the initial and final velocities, express the average velocity of the ball over time period [itex]t[/itex]. Next: Distance = average speed × time. You'll end up with a formula that starts with "[itex]s=...[/itex]". Finally, let [itex]s[/itex]=10 and solve for [itex]t[/itex].

This is really an unnecessarily long method, but by thinking several similar problems through in the same way you'll understand how to solve them. Next, derive for yourself the kinematics formulae listed in Georgepowell's post. Then you will have "earned the right" to cheat by using the formulas straight off the bat!

- m.e.t.a.