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My original answers that were both wrong:
a. 1230.436935
b. 57.2

hmm i cant get my brain into gear at the moment

haha, i appreciate your presence!

gotta go anyway im afraid

well did you use the tan ratio to find the horizontal distance..?

|dw:1442955509946:dw|

|dw:1442955391278:dw|
So my X velocity = 79.938
y velocity = 51.51645

in part (a) the speed of the plane has no relevance to the horizontal distance

Wouldn't it? Because the speed of the plan would be the initial velocity of the package?

\[x = 886/\tan(32.8)\]
So the plane traveled 1374.8 Meters

that's my answer for part (a)
someone will need to help you with part(b)

So for the package I got \[-886 = 51.516T - 4.9T^2\]
Trying to find the time it's in the air.

For the package T = 19.6945

i agree with 1574

Going for it!

Ugh stuck on
b. 29.37 Degrees didn't work.

No - I'm trying to remember how to find the angle)...

Hmm ok lemme roll with this for a minute... I THINK THAT'S IT!

the x vector would be 79.94 yes

no thats not right x is not 19.69 - thats t - the time

V = -51.51 + 9.8*19.69

that gives a velocity of 141.45 which seems feasible

so the required angle = arctan(141.45 / 79.94) = 60.53 degrees

So I got now it hits the ground at a 60.53 Degrees

yes
its been so long since i did these but i think that's correct

Well, I was very confident as well but still counting it wrong... I really dislike this WileyPlus

we both got the same answer but doing it different ways so ..

haven't done the actual numbers but it seems like most here are missing th epoint

did you check on 1574 - was that right?

That part was right! Sorry forgot to mention that. Only part b is wrong so far.

Distance the package travels in that time is the
x = 1574.33

hmm don't get - it I''m pretty sure you work out the angle using the velocity vectors..

at impact the Vx will be 79.94 and Vy - well it seems we did that correctly.

s = ut + 0.5 at^2
-881 = 51.5 t - 9.81/2 t
t=20

its -886

if your velocities are correct then
angle (wrt horizontal) = tan^-1 vy/vx

yes

anyway gotta go guys
its getting late here in the UK

Alright, thanks so much for the help!

have you been told a default value to use for g?
I used 9.8
time = 20 and vx = 80

Yes, I was using -9.8 as I was treating ground as 0.

I do appreciate the help.

I also re did the problem and got 60.5 degrees

I just don't see how that can be wrong...

and btw
horizontal distance travelled = vx * t = 1570m