Projectile, Finding Initial Velocity

[Laura1321412]

Homework Statement

A rocket is shot off a 3220 M high mountian at an angle of 35.3 to the horizontal. The rocket lands 9580 m away from the mountian, what was the initial velocity?

Homework Equations

y=Vi(y) *t +1/2gt^2

x= t*Vi(x)

Vi(y)= sin 35.3 Vi
Vi(x)= cos 35.3 Vi

The Attempt at a Solution

I rearanged the top equations to solve for t, then equated them, and replaced subbed the variables so the only var was Vi. When solving for Vi, i got some weird equations i don't know if I am using the right equations above, or what.

answers I've gotten: 318.7, 251.74

but neither are correct.

I don't know what to do! :(

 

[ehild]

Take care of the signs. If the vertical component of the initial velocity is positive the acceleration has to be negative.
How did you use the height of the mountain? ehild

 

[Laura1321412]

Here exactly what i did,,, sorry i should have written the whole thing out above.

time till rocket reaches highest point = Vi(y)/9.8
So, distance from top of mountian to highest point = Vi(y)/2 * Vi(y)/ 9.8 = Vi(y)^2/19.6

Then, distance from ground to highest point = Vi(y)^2/19.6 +3220
so, Vi(y)^2/19.6 +3220= Vi(0)*t + 1/2*9.8*t^2
- isolate t
sqrt(vi(y)^2/96.04 +657.14) = t time from highest point of flight to ground

So, total time for flight of rocket in y direction = sqrt(vi(y)^2/96.04 +657.14) + Vi(y)/9.8

And total flight time for x -> t = 9580/Vi(x)

Since the two t values have to be the same i equated the expressions

sqrt(vi(y)^2/96.04 +657.14) + Vi(y)/9.8 = 9580/Vi(x)

simplified a bit

vi(y)^2/96.04 +657.14 +vi^2/96.04 = 9580^2/Vi(x)^2

2Vi(x)^2Vi(y)^2/96.04 +657.14Vi(x)^2 = 9580^2

So, Vi(x)= cos 35.3Vi
Vi(y) = Sin 35.3 Vi

Sub these in

2(cos35.3Vi)^2(sin35.3Vi)^2/96.04 + 657.14(cos35.3Vi)^2= 9580^2

0.004632Vi^4 + 437.71Vi^2 = 9580^2

.0680Vi^2 + 20.92 Vi = 9580

Vi = 251.82


Checking this back with the values, this is too slow...

Im really confused! :S

 

[ehild]

So, total time for flight of rocket in y direction = sqrt(vi(y)^2/96.04 +657.14) + Vi(y)/9.8

And total flight time for x -> t = 9580/Vi(x)

Since the two t values have to be the same i equated the expressions

sqrt(vi(y)^2/96.04 +657.14) + Vi(y)/9.8 = 9580/Vi(x)

It is correct up to here, but the following is wrong.

simplified a bit

vi(y)^2/96.04 +657.14 +vi^2/96.04 = 9580^2/Vi(x)^2

(a+b)^2 is not a^2+b^2


ehild

 

[rwisz]

I would approach this problem by first identifying the given information and the unknown variable. In this case, the given information is the height of the mountain (3220 m), the angle at which the rocket was shot (35.3 degrees), and the distance the rocket landed from the mountain (9580 m). The unknown variable is the initial velocity of the rocket.

Next, I would use the kinematic equations to solve for the initial velocity. Since the rocket is launched at an angle, we can use the equations for projectile motion, which take into account both the vertical and horizontal components of the motion.

First, let's find the time it takes for the rocket to reach the ground. We can use the equation y=Vi(y) *t +1/2gt^2, where y is the vertical distance (3220 m) and Vi(y) is the initial vertical velocity (which we don't know). We can rearrange this equation to solve for t:

t = √(2y/g)

Plugging in the values, we get:

t = √(2*3220/9.8) = 25.04 seconds

Now let's find the initial horizontal velocity, Vi(x). We can use the equation x = t*Vi(x), where x is the horizontal distance (9580 m) and t is the time we just calculated. Again, we can rearrange this equation to solve for Vi(x):

Vi(x) = x/t

Plugging in the values, we get:

Vi(x) = 9580/25.04 = 382.5 m/s

Lastly, we can use the equations Vi(y) = sin 35.3 Vi and Vi(x) = cos 35.3 Vi to find the initial vertical and horizontal velocities, respectively. Plugging in the value we just calculated for Vi(x), we get:

Vi(y) = sin 35.3 * 382.5 = 219.2 m/s

Therefore, the initial velocity of the rocket is 382.5 m/s at an angle of 35.3 degrees to the horizontal.

In summary, to find the initial velocity of the rocket, we used the kinematic equations for projectile motion and solved for the unknown variables using the given information. It is important to carefully identify the given information and choose the correct equations to use in order to arrive at the correct answer.