/sci/ - Science & Math

[Anonymous No.4306797]

These are some pretty simple questions but I'm actually a retard can you help me out?
1. A 220 kg car is travelling at 45 km/h when its brakes are applied and it skids to a stop. If the coefficient of friction between the road and the tires is 0.70, how far does the car go before stopping?

2. Using a slingshot, you fire a stone horizontally from a tower that is 27m tall. It lands 122 m from the base of the tower. What was its initial velocity?

3. You throw a ball with a velcoty of 18 m/s at 24(degrees) above the horizontal from the top of your garage. 5.8 m above the ground. Calculate the
(a) time of flight
(b) horizontal range
(c) maximum height
(d) velocity when the ball is 2.0 m above the roof
(e) angle at which the ball hits the ground

[Anonymous No.4306801]

There is an equation for number 2. It's literally direct substitution.

[Anonymous No.4306803]

>>4306801
Oh, three too, actually. If anything, number 1's the only problem that requires some thought.

[cyriak No.4306804]

There's a formula for 1, but I dread doing anything other than browising 4chan with this archaic smartphone.

[Anonymous No.4306808]

>>4306801
o? what would the equation be?

[Anonymous No.4306824]

>>4306808
Try googling.

[Anonymous No.4306840]

>>4306824
I did, no luck

[Anonymous No.4306850]

>>4306840
I got a site that calculates that shit for you in 30 seconds. What kind of a retard are you?

[Anonymous No.4306855]

>>4306850
a holy retard...
what's the site?

[Anonymous No.4306983]

#2 is actually fairly simple, since there is no angle. The first thing you need to do is find time. using the equation: Y=Viy*T+.5*Ay*T^2 so, you know Y= -27 meters Viy=0 and Ay= -9.80 m/s so plug that in and you get= -27=(0)*T+.5(-9.80)*T^2 then simplify that since T*0=0 so the equation is now -27=.5*(-9.80)*T^2 then multiply what you know= -27=-4.9*T^2 divide=-5.51=T^2 square root and time= 2.35 sec.

Then take that and use the equation x= Vix*T+.5(Ax)*T^2, and You know X=122, T=2.35, Ax=0 so you can plug those in= 122=Vix*2.35+.5*0*2.35^2 and since that last part is times 0 you end up with 122= Vix*2.35 so divide and and you get Intial Velocity= 51.91

[Anonymous No.4307098]

>>4306983
wow, thanks a lot that looks like a hard one!

Well atleast someone helped...

[Anonymous No.4307250]

For the first one I arrive at s=v^2/(2*µ*g)
which means the mass doesn't matter
and you need (km/h)/3.6=(m/s)

[Anonymous No.4307680]

Sorry for taking so long after #2 had to eat and find the equations on how to do this one but first you break down your vector into its components using trigonometric functions so V=18 and Theta= 24 degrees so Vix= 18cos24=16.44 and Viy= 18sinc24= 7.32 keep in mind that Vfx is the same as Vix so we also know Vfx=16.44 we also have a Delta Y1=5.8.

With that we can calculate the Vfy using the equation Vfy^2= Viy^2-2gΔY so just plug in the numbers= Vfy^2=7.32^2-2(9.80)(-5.8)=167.26 then take the square root of that and you get Vfy=-12.93 negative because that component is in the -y direction

then since we have the 2 components we can find the magnitude and angle at which it hits magnitude is just Pythagorean Theorem so Vf=√Vfx^2+Vfy^2= √16.44^2+-12.93=20.92 then to find the angle at which it hit you use the inverse tan function so θ= tan^-1(Vfy/Vfx)= tan^-1(-12.93/16.44)=-38.18 degrees below horizontal

to find the time it takes use the equation Vfy= Viy-gt rearrange to solve for t so T= Vfy-Viy/g= T=(-12.93-7.32)/9.80= 2.07 seconds

once you know time you can find the displacement so ΔX= Xf- Xi=Xfx*t and Xi=0 so Xf= Xfx*t= 16.44*2.07=34.03 m

finally for maximum height use the equation Vfy^2= Vfi^2-2g(ΔY2) rearrang for Y2 and you get ΔY2=(Vfy^2-Vfi^2)/(-2(g)= (-12.93^2-0)/(-2(9.80))=8.53 the Vfi is 0 because at the top of the path the projectile has slowed down its y velocity to 0 and will start to fall again.