# Thread: Physics: Magnitude of initial velocity

1. ## Physics: Magnitude of initial velocity

Problem:
A frog makes a mighty leap, leaving the ground at an angle of 45 degrees and traveling a horizontal distance of 1.47 (m) in 0.55 (s).

a)What is vx?
b)What is vyi?
c)What is the magnitude of the initial velocity, at an angle of 45 degrees?

Now I solved question b by drawing a little figure and finding vyi, and used that to find time and then used that to find vx. But im not quite sure how to find the magnitude?

Δy=?
vyi=2.67 (m/s)
vyf=?
a=-9.8 (m/s^2)
t=.55 (s)

Δx=1.47 (m)
vx=2.76 (m/s)
t=.55 (s)

2. a)What is vx?
I got 2.67 instead of 2.76, but I am guesssing that that's just a typo.

b)What is vyi?
I got a different answer to yours. We have
a = -9.8
Δy=0
t=.55 (s)
vyi=?
Since we don't care about final velocity, use the equation $\displaystyle \delta y = v_{yi}t+0.5at^2$ and solve for $\displaystyle v_{yi}$

c)What is the magnitude of the initial velocity, at an angle of 45 degrees?
magnitudes of vectors can be found using triangles like this one. You know $\displaystyle \theta = 45^o$ and $\displaystyle v_{yi}$. You want $\displaystyle |V_i|$

3. Thanks!
That helped a lot

I got the magnitude as 3.77 (m/s). Is this correct?

Another question I have is on a different problem similar to this one and they want to know the direction, and I have no idea how to find that?

4. I got the magnitude as 3.77 (m/s). Is this correct?
No. But I think it was my fault: the formula for viy I posted before was wrong and has now been fixed. You should get 2.95 for viy and about 3.8113055505954911562055111 for the magnitude

Finding theta in the triangle above gives direction

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### how to find magnitude of initial velocity formula

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