Question

A volcanic block is ejected at an angle of 45 degrees from Mount Fuji during a volcanic eruption. It lands at the foot of the volcano at an horizontal distance of 9.4 km. The height of Mount Fuji is 3.3 km. What is the block's initial speed?

Do I use Pythagorean theorem to find the hypotenuse?
Is my distance known?

Answer 1

I got the formula x=x(initial)=v(inital x)*t+1/2at^2

Answer 2

Nope you would use this formula
\[R=\frac{ v _{i}^2\sin2\theta }{ g }\]
where R is the horizontal distance it traveled
vi is intial velocity
theta is the angle
g is the acceleration of gravity

Answer 3

Your formula is incorrect because it is a 2 dimensional motion (x and y component changes)

You can ignore the height of the volcano

Answer 4

But to find R convert the horizontal distance of km to meters

Answer 5

The think is I never learned that formula/ This is a calc base physic. So I need to to get the velocity by figuring out what I have. I drew a triangle with the angle

I get \[V _{0x}= v _{0}*\cos(45)\]
\[V _{0y}= v _{0}*\sin(45)\]

Answer 6

besides I don't know what's R and the initial velocity

Answer 7

You would need initial velocity to use that formula of yours.
It would be the x and y component of the initial velocity.
Where were you given "that" in the question?

Answer 8

there is not a initial velocity. This is suppose to make you find it. So I was thinking. when t(time) is 1/2 does that equal 0 for the y component since it is tangible half way making no velocity when time reaches 1 half of it's destination?

Answer 9

\[v _{xi}=v _{i}\cos \theta \]
\[v _{yi}=v _{i}\sin \theta \]

Answer 10

I did that. But what V initial?"

Answer 11

Thats what I said.
You mentioned the formula I wrote and I told you that it wasnt needed.

Answer 12

But I dont think my professor wants to see that since I have not learned it yet. He want's me to use what I have which is the distance formula and velocity/

Answer 13

so the v( y initial) is irrelevant?

Answer 14

You sure?
Check your notes or online lessons?

Answer 15

yes.
for a thrown object,
a=g
v=gt+v
x=1/2 t^2 + v(initial)

Answer 16

Its a projectile motion.

It is 2D dimensional