# How to find initial velocity without being given time

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1. Feb 10, 2015

### Qwerty123!

1. The problem statement, all variables and given/known data
a projectile is launched at an angle of 26.5 above the horizontal at a height of 1.11m. the range is 2.331m. find the initial velocity and time given that acceleration due to gravity is 9.8m/s^2.

3. The attempt at a solution
first, I split the velocity into its components:

v1x = v1cos26.5
v1y = v1cos26.5

then,
v1x = dx/t
v1cos26.5 = dx/t
rearranged to isolate for time since it is not given:
t = dx/ v1cos26.5

also, using the kinematic equation: v2y^2 = v1y^2 + 2at
and since at the projectile's max, v2y = 0m/s and since the max occurs at t*(0.5) ***(but I think this is inaccurate because the projectile does not land at the same height in which it was released)

so,
v2y^2 = v1y^2 + 2at
0 = v1y^2 + 2(9.8)(t/2)

then isolate for t again,

t = 2(v1y)^2 / 4.9
or
t = 2(v1sin26.5)^2 / 4.9

then make each isolated time equation equal one another:

dx/v1cos26.5 = 2(v1sin26.5)^2 / 4.9

then isolate for v1:
v1 = sqrt( 4.9*dx / (sin26.5cos26.5) )
since dx = 2.331,
v1 = sqrt( 4.9*2.331/ (sin26.5cos26.5) )
v1= 5.35m/s

Therefore, v1 = 5.35m/s

then for time,
v1x = dx/t
t = dx/v1cos26.5
= 2.331 / 5.35cos(26.5)
= 0.487s

Therefore, t = 0.49s

2. Feb 10, 2015

### Staff: Mentor

Hi Qwerty123!. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

The correct equation is v2 = u2 + 2as :(

See how you go now.

Last edited by a moderator: May 7, 2017
3. Feb 11, 2015

### Qwerty123!

Thanks for replying. :). I'm not familiar with the variables u and s?

4. Feb 11, 2015

### Staff: Mentor

It's a common convention to let u be the initial velocity and s represent distance ("s"pace I suppose).

5. Feb 11, 2015

### Staff: Mentor

Maybe you'd like it as: v2y^2 = v1y^2 + 2.a.d
where that last term involves the y component of acceleration times distance in y direction.