anonymous
  • anonymous
Two seconds after being projected from ground level, a projectile is displace 40mhorizontally and 53mvertically above its launching point. What are the (a)horizontal and (b)vertical components of the initial velocity of the projectile? (c)At the instant when the projectile achieves its maximum height above ground level, how far is it displace horizontally from the launch point.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
horizontal displacement is constant, vertical is dependant on r(t)=vt-5t^2 (s=s'+ at 2 secs we get r(2)=2v-5*4=53 v=(53+20)/2 v=36.5m/s this is the vertical component horizontally we get that r(t)=vt r(2)=2v=40 v=20m/s to solve the last you could either apply x(axis) or the derivative
anonymous
  • anonymous
d/dt r(t) = v-10t 36.5-10t=0 t=3.65
anonymous
  • anonymous
horizontal its 26.5m/s

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anonymous
  • anonymous
sorry i mean vertical lol
anonymous
  • anonymous
dunno man.. i still think it is 36.5 r(2)=v*2-5*4=53 53=2*v-20 2v=73 v=36.5
anonymous
  • anonymous
this is more of a physics question
anonymous
  • anonymous
oh, to the one with the question, anything u did not understand feel free to ask dude!
anonymous
  • anonymous
isn't it 53m/2s 26.5/s
anonymous
  • anonymous
forgot to add the deacceleration! gravity influencing over the vertical speed
anonymous
  • anonymous
so v=v0+at 0=26.5-9.8t t=2.7s
anonymous
  • anonymous
its asking to find the velocity which is the displacement
anonymous
  • anonymous
u using the constant motion formula, but the projectile is submitted to gravity
anonymous
  • anonymous
so acceleration of gravity only affects y when your looking for the y value?
anonymous
  • anonymous
so when do you use v=v0+at and when do you use x-xo=v0t+1/2at^2?
anonymous
  • anonymous
|dw:1333234262419:dw| vertical: r(t)=r0+v0*t-1/2*g*t^2 horizontal: r(t)=r0+v0*t
anonymous
  • anonymous
oh i see now lol damn
anonymous
  • anonymous
u use the first one when there is no acceleration influencing the system. the second one is obtained when u get a constant acceleration.
anonymous
  • anonymous
but there is gravity in the first formula
anonymous
  • anonymous
or did you mean the other formula
anonymous
  • anonymous
yeah g=10 so u have the r0=0, v0=?, g=10 r(t)=v0*t-1/2*10*t^2 => r(t)=v0*t-5*t^2 plug in 2 seconds and u can find v0
anonymous
  • anonymous
note that I came to the same expression I used in the first place by plugging in g=10 the acceleration in the system is g
anonymous
  • anonymous
ok found y v0 but don't get when you use the formula v=vo+at...is it normally for horizontal motion?
anonymous
  • anonymous
i mean in general
anonymous
  • anonymous
thing is, the projectile is submitted to only one velocity. however, we can trace its projections in X and in Y, respectively the horizontal and the Vertical components of the displacement. analysing the system u see that there is acceleration only in the y axis (gravity in this case) generally, use the more complex formula to movements submitted to acceleration sorry about my aweful english, I am brazillian and only took those lessons in portuguese
anonymous
  • anonymous
awful*
anonymous
  • anonymous
lol ah k but then when do you use v=v0+at? is it when there is no distance measurements given?
anonymous
  • anonymous
this is the equation to find the velocity over time, note that the v is in funtion of a and t. guess u are in doubt where u use r(t)=r0+vt? this is the equation when there is no acceleration(roughtly r(t)=r0+v0t+a+t^2*1/2 where a=0)
anonymous
  • anonymous
ah k so damn its confusing...ah well damn book
anonymous
  • anonymous
thanks
anonymous
  • anonymous
feel free to ask anything else dude!! u need to give urself some time to think about it, it doesn't come so natural overnight, give urself some time! no stress
phi
  • phi
so you have the initial horizontal velocity as Vx= 20 m/s using Sx= Vx * t with t=2 sec and Sx= 40m you have the initial vertical velocity Vy= 36.5 using Sy = Vy*t - 0.5 g t^2 = 2 Vy - 5*4 = 53 m. solve for Vy for part (3) use the initial vertical velocity Vy= 36.5 m/s and find when it has slowed by gravity to 0 (where it will be at the max height). Vy = g t, 36.5/10 = t, t= 3.65 sec use Vx * t = S, to find 20 m/s * 3.65 sec = 146 m horizontal displacement.

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