Discern initial V w/o components?

  • Thread starter Thread starter afronicus
  • Start date Start date
  • Tags Tags
    Components Initial
AI Thread Summary
To find the initial velocity (Vi) of a ball that traveled 188 meters before landing, the discussion emphasizes breaking Vi into its x and y components due to the 45° launch angle. The horizontal motion indicates that the time of flight can be expressed as t = 188 / (Vi cos 45°). The vertical motion must account for the initial height of 0.9 meters and the acceleration due to gravity (9.8 m/s²). By substituting the time into the vertical motion equations, the participants aim to solve for Vi, ultimately leading to a numerical value of approximately 41.3 m/s. The conversation highlights the importance of correctly manipulating the equations to isolate Vi.
afronicus
Messages
3
Reaction score
0
Discern initial V w/o components!?

Homework Statement



Ok, ball traveled 188m (618 ft) before landing on the ground. Assume ground is flat, and ball Vi is from 45o, disregard air resistance, and Starting position is .9m above ground.
Find Vi

I KNOW )))) a= 9.8 m/s2
Xi= 0
Xf=183
Yi= 0.9m
ANGLE= 450
Vy (@ highest point) = 0



Homework Equations


Ok so the equations I'm using are:
A---Vf= Vi + (a)(t)

B---Xf= Xi+ Vi+ (1/2)(a)(t)2

C---(Vf)2=(Vi)2+2a(Xf-Xi)

D---Xf-Xi=((Vi+Vf)/2)(t)

The Attempt at a Solution



So I've been racking my brain over this for the past hour, and I keep failing to discover the answer. My equations are left with 2 variables. What am I missing?? :/
 
Physics news on Phys.org


afronicus said:

Homework Statement



Ok, ball traveled 188m (618 ft) before landing on the ground. Assume ground is flat, and ball Vi is from 45o, disregard air resistance, and Starting position is .9m above ground.
Find Vi

I think you've overloaded your brain with information and equations, so let's take it slowly.

Split the initial velocity Vi acting at 45°, into x and y components.

The first line says
afronicus said:
ball traveled 188m (618 ft) before landing on the ground.

So this means it traveled horizontally 188 m. So what does this tell you about the time taken for the motion? (Hint: horizontal velocity remains constant throughout, so distance = speed*time)

Write the time,t, in terms of Vi.

When you get this, then we will consider vertical motion.
 


Vx=x/t so that will help you.
 


hi, thanks a bunch for your reply. Ok, so my Vi in components is Vix=Vi cos(45)
&
Viy=Visin(45)
...
Ok, and time can be written as:

t =( Vfx/y / a) + Vix/y.

(Vx)(t)=188 m

apparently there is numerical value you're supposed to get...I opted to have Mastering Physics show me the answer, said it was 41.3. I just can't figure out how.
...
 


A man, my formula B is off, initial veloss should be multiplied by time. Thought I had figured out out... Not so much.
 


afronicus said:
(Vx)(t)=188 m

apparently there is numerical value you're supposed to get...I opted to have Mastering Physics show me the answer, said it was 41.3. I just can't figure out how.
...


afronicus said:
A man, my formula B is off, initial veloss should be multiplied by time. Thought I had figured out out... Not so much.

So your time would be

t = \frac{188}{V_i cos 45}

I can't really read how you manipulated formula B, but considering y-direction, the initial displacement y0 you were given. And you know that y = 0 when t = 188/(Vicos45°).

Just plug it in and solve for Vi.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top