Can someone please please help me with these physics problems. I don't really know how to do them and I have

Question

to take my semester exam tomorrow over these kind of problems.


A rock is projected from the edge of the top of
a building with an initial velocity of 23:8 m=s
at an angle of 30± above the horizontal. The
rock strikes the ground a horizontal distance
of 96 m from the base of the building.
The acceleration of gravity is 9:8 m=s2 :
Assume: The ground is level and that the
side of the building is vertical. Neglect air
friction.
What is the horizontal component of the
rock's velocity when it strikes the ground?
Answer in units of m=s.

How long does the rock remain airborne? An-
swer in units of s.

How tall is the building? Answer in units of
m.

What is the vertical component of the rock's
velocity when it strikes the ground? Answer
in units of m=s.

What is the magnitude of the rock's velocity
when it strikes the ground? Answer in units
of m=s.


A cannon sends a projectile towards a target
a distance 1480 m away. The initial velocity
makes an angle 28± with the horizontal. The
target is hit.
The acceleration of gravity is 9:8 m=s2 :
What is the magnitude of the initial veloc-
ity? Answer in units of m=s.

How high is the highest point of the trajectory? Answer in units of m.

How long does it take for the projectile to
reach the target? (Assume no friction) Answer in units of s.

A cat chases a mouse across a 1.0 m high
table. The mouse steps out of the way, and
the cat slides o® the table and strikes the °oor
2.4 m from the edge of the table.
The acceleration of gravity is 9:81 m=s2 :
What was the cat's speed when it slid o®
the table? Answer in units of m=s.

A daredevil decides to jump a canyon. Its
walls are equally high and 14 m apart. He
takes o® by driving a motorcycle up a short
ramp sloped at an angle of 12±.
The acceleration of gravity is 9:8 m=s2 :
What minimum speed must he have in order to clear the canyon? Answer in units of
m=s.

Answer 1

These questions center around using trigonometry to separate the for the vertical velocity components that are affected by gravity from the horizontal components that are not.

And it's pretty simple stuff. 90% of trig is the 4 equation devoted to finding how much shorter it is to cut across a field than to walk along its edges.

You are standing on the South-West corner of a square field. Now to get anywhere on the East side you can walk East than North, OR you can pick an angle & cut across the field. The distance you would walk East is called the adjacent side, because it's adjacent to the angle of the short cut. The distance you'd walk North is called the Opposite side because it's opposite to the angle. And the distance of the short cut is called the Hypotenuse just to confuse you. ;-)

Here's the deal

H^2 = A^2 + O^2
Sin (angle) = O/H
Cos (angle) = A/H
Tan (angle) = O/A

And that's pretty much it. The sum of square of the sides equals the square of the shortcut. And if you have a calculator & know the angle, you can find any ratio between any of the sides; which means if you know one side you can find any other.

Now if you separate the total velocity of your rock (H) into it's Horizontal (A) & Vertical (O) componants, you problem becomes much easier to solve.

Your horizontal component (A) would be H * A/H - for the angle in question.
I.e. 23.8 m/s * Cos (30)
Got it?
Your vertical component (O) would be H * O/H or 23.8 m/s * Sin (30)

So if you know horizontal velocity and how far your rock traveled horizontally (96m) you should be able to calculate the time it spent in the air. Right.

If you know the vertical velocity you should be able to calculate how high & how long it could fly up before it stopped. Right?

Subtract this from the total time in the air & you know how much time it spent falling. Which means you should be able to calculate how far it fell. Subtract how far it rose from how far it fell and you have the building height.