physics homework?

Question

an object dropped from the top of a cliff takes 1.80 seconds to hit the water in the lake below, what is the height of the cliff above the water?

if a dropped object falls 19.6 meters in 2.00 seconds, how far will it fall in 4.00 seconds?

a ball is dropped out of a window near the top of a building. if it accelerates towards the ground at a rate of 9.80 meters per square seconds, what is its velocity when it has fallen 4.00 meters?

a boy on a bridge throws a stone vertically downward toward the river below with an initial velocity of 14.7 meters per second. if the stone hits the water 2.00 seconds later, what is the height of the bridge above the water?

Answer 1

the formula you can use is

s = a*t*t + v*t + h

where
s is the traveled distance,
a is the acceleration usually 9.8
v the initial velocity
h the already travveled distance
t time in seconds

for instance the first one
t = 1.8 seonds
a = 9.8 meter per seconds
YOU need to calulate the s.
all other variables are not given thus 0.
s = 9.8*1.8*1.8 + 0*1.8 + 0 = ... use your calculator.

Answer 2

These problems all rely on the same basic principle - applying the equations of motion for uniform acceleration.

I won't supply the answers, since this is your homework, but i'll give you step-by step solutions for the first and third ones, and hopefully you'll get the idea.

Let's first write down all five variables for the first one
u = initial speed = 0ms^-1 (it's DROPPED)
v = final speed = ? (we don't know it)
a = acceleration = g = 9.8ms^-2 (the acceleration due to gravity on the earth's surface - assume constant)
s = height = what we want to find
t = time taken = 1.8 seconds.

now we've written down all the information we have in an easily accessible form.
Now you just have to pick the easiest equation of motion to use from
v = u + at
s = ut + 1/2at^2
v^2 - u^2 = 2as
s = (u + v) * t / 2

In this case we know u, a and t
so we need to use s = ut + 1/2at^2 because it contains only those variables that we know plus the one we want to find.

Substitute the numbers into the formula and you get the height.



Now let's look at number 3, because it's slightly different. This time you want to know the speed when it's fallen 4m.
here we go:
u = 0ms^-1 (dropped)
v = ? = what we want to find
a = g = 9.8ms^-2
s = 4m (how far it's fallen at the point of interest)
t = don't know and don't care

now you simply pick the correct equation of motion again.
This time it's v = u + at.

As you can see, these problems only require that you write down all the information you are given in an easy-to-read form, and choose the equation which possesses all the information you know plus what you want to find. Then just rearrange for what you want (if necessary) and plug the numbers in and you get it!

Hope this helps! If you want more detailed information about something just ask.

Answer 3

Someone who actually admits they are asking for help with homework *gasp*. Just for that, you get actual help.

The cliff: Integral of velocity. You've seen s=ut + 1/2 at^2?
Here; u=0, I'm assuming it's on Earth, so a=9.8, t=1.8.
s=0 + 1/2*9.8*1.8^2= 15.876 m.

The object that fell 19.6 m in 2s. Use the same formula, just rearranged. s=19.6, u still= 0, t=2. 19.6=0 + 1/2a*2^2.
a=9.8, so we're probably still on Earth. Then just put that back in,
s = 0 + 1/2*9.8*4^2 = 78.4 m.

Ball out of window, for this one you use v^2 = u^2 + 2as. As always, u=0, a=9.8, s=4. v^2=0+2*9.8*4=78.4, v=8.854ms^-1.

Boy on bridge: back to s=ut+1/2at^2. This time u=14.7, a=9.8, t=2. s=14.7*2+1/2*9.8*2^2=49m.

If only I was back in high school with such easy questions...

Answer 4

1) Gravity = 9.8ms, if takes 1.8 seconds means 1.8 * 9.8 = 17.64m

2) gravity = 9.8ms
9.8 / A = 19.6 / 2, there for A = 1 means the 4 s = 4 * 9.8 = 39.2

3) it should be constant at gravity speed =9.8ms

4) 14.7 x 2 = 29.4

Answer 5

A)
The object dropped starts at 0-velocity:

0 = x0 + 0*(1.8) - (1/2)(9.8)(1.8^2)
x0 = (1/2)(9.8)(1.8^2) = 15.88 meters

B)
19.6 = (1/2)(9.8)(2^2)
x = (1/2)(9.8)(4^2)
=78.4 meters

C)
(v2)^2 = (v1)^2 + 2ad
(v2)^2 = 0 + 2(9.8)(4) = 78.4 (m/s)^2
v2 = 8.85 m/s

D)
s = (14.7)(2) + (1/2)(9.8)(2^2)
=49 meters

Answer 6

Yes, that's physics homework. I suggest you use the constant acceleration equations.