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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
d/dt r(t) = v-10t 36.5-10t=0 t=3.65
horizontal its 26.5m/s
sorry i mean vertical lol
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
this is more of a physics question
oh, to the one with the question, anything u did not understand feel free to ask dude!
isn't it 53m/2s 26.5/s
forgot to add the deacceleration! gravity influencing over the vertical speed
so v=v0+at 0=26.5-9.8t t=2.7s
its asking to find the velocity which is the displacement
u using the constant motion formula, but the projectile is submitted to gravity
so acceleration of gravity only affects y when your looking for the y value?
so when do you use v=v0+at and when do you use x-xo=v0t+1/2at^2?
|dw:1333234262419:dw| vertical: r(t)=r0+v0*t-1/2*g*t^2 horizontal: r(t)=r0+v0*t
oh i see now lol damn
u use the first one when there is no acceleration influencing the system. the second one is obtained when u get a constant acceleration.
but there is gravity in the first formula
or did you mean the other formula
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
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
ok found y v0 but don't get when you use the formula v=vo+at...is it normally for horizontal motion?
i mean in general
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
lol ah k but then when do you use v=v0+at? is it when there is no distance measurements given?
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)
ah k so damn its confusing...ah well damn book
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
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.