How Do You Calculate Initial Velocity Components Given Angle and Distance?

[gsg822]

Homework Statement

I need help finding the x and y components of initial velocity being given the angle the trajectory was fired at and the distance the projectile traveled. For this example I'm using an angle of 15° and the projectile travels 0.534m. I'm not sure if I'm going in the right direction in trying to solve this problem.

Homework Equations

Vx = Vcos(angle)
Vy = Vsin(angle)

x -x initial = Vx * t

The Attempt at a Solution

Solved those to get: Vx = Vcos(15) and Vy = Vsin(15).

After pluggin Vx into the equation, I'm having trouble figuring out where to go from here. Any help is appreciated.

 

[tiny-tim]

welcome to pf!

hi gsg822! welcome to pf!

(try using the X₂ button just above the Reply box )

… I'm using an angle of 15° and the projectile travels 0.534m.
…
Solved those to get: Vx = Vcos(15) and Vy = Vsin(15).

After pluggin Vx into the equation, I'm having trouble figuring out where to go from here. Any help is appreciated.

use the x equation to find t (as a function of V), then use that value of t in the y equation …

show us what you get ​

 

[gsg822]

When you say to use the y equation are you referring to y = y₀ + v_oy * t - 0.5 * g * t². I've tried using this equation and I get stuck when I try and simplify it down.

I solved the x equation for time and got t = 0.534m / (V * sin(15))

 

[tiny-tim]

I solved the x equation for time and got t = 0.534m / (V * sin(15))

yup!

When you say to use the y equation are you referring to y = y₀ + v_oy * t - 0.5 * g * t². I've tried using this equation and I get stuck when I try and simplify it down.

if I'm reading the question correctly, y = y_o

 

[gsg822]

I have two variables, V and V_oy.

0 = V_oy(0.534m / Vsin(15)) - (1/2)(9.8m/s²)(0.534m / Vsin(15))²

(1/2)(9.8m/s²)(0.285m / (Vsin(15)²) = V_oy(0.534m / Vsin(15))

2.79m²/s² / ( Vsin(15)²) = 2V_oy(0.534m / Vsin(15))

2.79m²/s² / ( Vsin(15)) = 2V_oy(0.534m)

I get stuck here and don't know where to take it from here. I'm also not really sure if this work I did is even right. Am I heading in the right direction?

 

[tiny-tim]

V_oy = Vcos15°

(and you could have divided the whole of that equation by t, since you had 0 on the LHS)

and now I'm off to bed :zzz:​

 

[gsg822]

I forgot that I could substitue Vsin(15) for V_oy.

I think I may have gotten the answer.

Vsin(15) = 5.2 / 2Vcos(15)

Rearranged that to V² = 5.2 / (2sin(15)cos(15))

V = sqrt(5.2 / (2sin(15)cos(15)))

V = 3.22 m/s

Plug that value back into the V_ox and V_oy equations to get:

V_ox = 3.11 /s
V_oy = 0.833 m/s

Could you verify if this is correct?

 

[tiny-tim]

hi gsg822!

(i'm sorry for the delay )

where did this come from? …

Vsin(15) = 5.2 / 2Vcos(15)