If you help with my physics homework! Yes, more physics! YAY! 2D Kinematics, 9.8 gravity [as opposed to 9.81]:

ReT is pushed off of a 78.2 cliff at 13.5 m/s. How far will he land off the base of the cliff?

78.2 cliff at 13.5 m/s. How far will he land off the base of the cliff?

look up the equation first, I can't remember it off the top of my head

Find out how long it'll take him to hit the bottom accelerating at 9.8.

Then plug that time into the speed that he's moving sideways.

Yeah, but how do I find time? My equation sheet says:
(V[yi] is Initial Y velocity.)

h[subscript]p = (V[yi]^2) / 2|g|

If h[subscript]f == 0:

t = (2V[yi]^2)/(2|g|)

If h[subscript]f != 0:

t = (V[yi] +/- SQRT(V[yi]^2-2|g|h)) / |g|

What the HELL is h[subscript]f and h[subscript]p?

You have the height (which is in meters)
you have the acceleration ( meters per second squared)
you have a speed (meters per second)

There is definatley a way to take what you have there and find the time.

Forget doing this with physics and equations.

The best way is to literally push him off of a cliff and see how far away he lands. :p

Distance traveled = initial velocity(time) + 1/2 acceleration(time ^2)

Solution!

I used the time equation for H[f] == 0 and got 2.76 for time.
When I use the kinematics equation, I get 18.63 m

But JG gets 53.9. He used V[x] = V*cos(theta). He also used 3.992 for time.

Who's right?

This reminds me of an old IRC quote.

* Flirbnic pushes Genki down the hill again.
[17:30:41] * Genki misses the rocks and falls off the cliff
[17:36:34] * Genki lands
<Flirbnic> Wow
<Flirbnic> That must've been a pretty tall cliff...
<Genki> sure
<Flirbnic> 611.207 kilometres, in fact
<Flirbnic> And you impacted on the ground at approximately 3.463 km/sec
<Flirbnic> That's more than ten times the speed of sound.
* Genki (O_o@holonet-17A0B186.mm-interactive.com) Quit (Ping timeout)

ReT will hit the ground at 54 meters from the cliff, in an interval of 4 seconds.

Why does my time differ from yours?

Here's what I used to find the time:

distance = initial velocity * time + 1/2 * acceleration * time^2
78.2 m = 0 m/s * time + 1/2 * 9.8 * time^2

Time = 3.99 sec

Yeah, I figured it out now.

Next question:

An object is thrown off of a 75.4 m cliff. The ball lands 155 m from the edge of the cliff. What is the initial horizontal velocity?

(V[x] means Horizontal Velocity)

D = V[x]*t

I need to find time...

Just remember, vertical velocity does not affect horizontal velocity.

Find time the same way you did in the last problem.

Can't. We don't know initial velocity like we did in the last equation.

>.> landed on my feet

I get 39.5 m/s.

Next question:

A cheerleader is cheering on the basketball team when her shoe comes off. If she kicks the shoe with an initial velocity of 14.2 m/s at an angle of 55.8, how high will her shoe go?

Subsequently, this cheerleader sometimes misses practice for cheerleading :-P

14.2 m/s * sin(55.8) = 11.7 m/s;

Ek = (1/2)mv^2; Ep = mgh;
(1/2)mv^2 = mgh;
(1/2)v^2 = gh;
h = 7.0 m

Exactly what I got.

Wel, 6.89, which is more precise.

Oh. If you're showing your work you can't do it the way I did. I don't think you're quite there yet and your teacher will know what's up.

Edit: Also, remember your significant digits. Your gravitational acceleration constant is 9.8 m/s^2. And you should also remember the difference between precision and accuracy.

Next:

Tiger Woods hits a golf ball with an initial velocity of 43.2 m/s at an angle of 30 degrees. How far from the tee does the ball land? (The ball lands at the same height as the T)

d = Vx * t

Vx = 43.2*cos(30) = 37.4
t = (2(Vyi))/|g|
Vyi = 43.2*sin(30) = 21.6
t = (2(21.6))/|g| = 4.41

d = 37.4 * 4.41 = 165m

And then: A soccer ball is kicked with an initial velocity of 21.6m/s at an angle of 32 degrees. How long is the ball in the air?

t = SQRT(2h/g)

h = Vyi^2/2|g|

Vyi = 21.6 * sin(32) = 11.4

h = 11.4^2/2|g| = 129.96/19.6 = 6.62

t = SQRT(2*6.62/g) = 1.35s

[QUOTE=Lord Kuat]You know, you should really do these on your own... It will be more helpful to you that way.[/QUOTE]

I just did 2 on my own! Telling me if I did it right will help me even MORE.

Bah, I deleted my post too slowly! Yeah, I was reponding to the initial one, then saw later on. Campers, lolz...

Kinematics was the easy part of Physics.

Actually you do. In this senerio, the initial VERTICAL velocity is 0m/s since the object was thrown horizontally (It didn't exactly say that, word for word, in the question, but that is what the question implies. It's impossible any other way too, unless you know V0). Always remember, the vertical and horizontal velocities use separate equations. You can use the vertical velocity to find the time, then use the time to find the horizontal velocity. If you still don't understand, I'll write out the math, I'm just lazy right now.

[edit] Pommy pointed out that there could be a misunderstanding here, I fixed it. [/edit]

Come on, dude. These are ridiculously easy. You really don't need our help.

Anyways, for this one

A soccer ball is kicked with an initial velocity of 21.6m/s at an angle of 32 degrees. How long is the ball in the air?


0 = 21.6 sin 32* + (-9.8t1)

t1 ~= 1.168 sec

therefore t ~= 2.336 sec



What you did wrong:
1) your t= is for time for soccer ball to reach apex (when v=0). Well, firstly, you forgot to sqrt it. If you had, you wouldve gotten 1.16, which is what I got for half time. Then, multiply it by 2 to get the full time in air.

You could've saved a lot of grief just by setting y-velocity = 0 (because when ball is at apex y-velocity is at 0 (BUT OVERALL VELOCITY IS NOT)) and using vf = v0 + at (see my work above)

Just a clarification, not always always 0, but rather always 0 when initial velocity is only horizontal (see sg-fan's explanation)


Tiger Woods hits a golf ball with an initial velocity of 43.2 m/s at an angle of 30 degrees. How far from the tee does the ball land? (The ball lands at the same height as the T)

x = vt + at^2


0 = 43.2 sin 30* + (-9.8t1)

t1 ~= 2.204
t ~=4.408

x ~= (43.2 cos 30*)4.408 + (0)
x ~= 164.913

You got that one right