SRL Archive - Mathematics

I have a test in a few mins and I really really need a quick answer <3

the question is:

What is the approximate distance a baseball will travel if you throw it at a velocity of 97 ft/s and at an angle 30 degrees level to the ground?

E: I don't mean the distance along the curve, but from starting point to end point

E2: There's another question I need to know:

What are the components of a vector in which an airplane is traveling 480 mph at an angle of 280 degrees when there is wind pushing from 265 degrees?

The range of a projectile is the speed squared divided by g times the sin of 2 times theta.

d = (v^2/g)*sin(2*theta)

From: tehq
The range of a projectile is the speed squared divided by g times the sin of 2 times theta.

d = (v^2/g)*sin(2*theta)

I can't exactly say this helped a lot

i'm supposed to use an equation close to:
s= -16t^2 + initial velocity (t) + height above ground

but i don't know how to add the angle in

The answer to the first one should be 254 ft.

e1: The answer to the second one is conditional of your knowledge. To me, the answer would be a thrust forward (figure out angle), a drag backwards, a lift directly upwards, the weight directly downwards, and the relative wind backwards.

From: tehq
The answer to the first one should be 254 ft.

Thanks <3
And if you could get the 2nd one… that would be AMAZING

From: Bobby Boo

What are the components of a vector in which an airplane is traveling 480 mph at an angle of 280 degrees when there is wind pushing from 265 degrees?

280 degrees? Usually 0 degrees = left, so 270 degrees would mean it's like uhm crashing into the ground.
If so, vector has X and Y component.
X-component = cos(280degrees)*480
Y-component = sin(280degrees)*480

Same with wind,
X-component = cos(265degrees)*windspeed
Y-component = sin(265degrees)*windspeed

From: Markus
280 degrees? Usually 0 degrees = left, so 270 degrees would mean it's like uhm crashing into the ground.
If so, vector has X and Y component.
X-component = cos(280degrees)*480
Y-component = sin(280degrees)*480

Same with wind,
X-component = cos(265degrees)*windspeed
Y-component = sin(265degrees)*windspeed

Perhaps the plane is taking off upside down, or 280 degrees was a typo of 180? I'm fairly certain 280 degrees would be an almost dangerous pitch for a plane! (It would be upside down and accelerating extremely fast toward the ground at least assuming that it started at zero and rotated around counter clockwise).

From: Markus
280 degrees? Usually 0 degrees = left

Just letting you know that the standard convention for this is that 0 degrees is normally taken as to the right then working anti clockwise from here: #****** 360 should be 270 woops

(atleast thats how us engineers are taught it in the UK)

______

For the first one just use your equations of motion:


For the second one I would draw a free body diagram of the plane and all the forces mentioned acting on it. Then resolve all these forces into only vertical and horizontal forces. now that you have them using Pythagoras' theorem work out the resultant force and the direction in which it acts.
(I'm pretty sure this is what the question meant) EDIT: NVM its late just realised I didn't even read the question right and I jumped ahead of myself in thinking it was something else

E2: going through the first one and I didn't get the same answer as tehq :/

I used (V-U)/a=t in order to find the total time the ball spends in the air as 3.014s
now that I know this value I used S=((V+U)*t)/2 to find the distance travelled.
obviously each time I used the corresponding values of V and U since the first equation is in terms of vertical motion and the second one is in terms of horizontal motion.
I got my answer as 35.587m (yes I use a real measurement system not dumb ft) which is 116.755249 feet