Answer:
[tex]v_0[/tex]=4.761 m/s
t=0.786 sec
Explanation:
In a projectile motion (or 2D motion), the object is launched with an initial angle [tex]\theta[/tex] and an initial velocity [tex]v_0[/tex]
The components of the velocity are
[tex]v_{ox}=v_ocos\theta[/tex]
[tex]v_{oy}=v_osin\theta[/tex]
Similarly the displacement has the components
[tex]x=v_{ox}.t=v_ocos\theta.t[/tex]
[tex]y=v_osin\theta.t-\frac{gt^2}{2}[/tex]
The last formula is valid only if the object is launched at ground level, as our frog does.
There are two times where the value of y is zero, when t=0 (at launching time) and when it lands back from the air. We need to find that time t by making y=0
[tex]0=v_osin\theta.t-\frac{gt^2}{2}[/tex]
Dividing by t (assuming t different from zero)
[tex]0=v_osin\theta-\frac{gt}{2}[/tex]
Then we find the total flight as
[tex]t=\frac{2v_osin\theta}{g}[/tex]
Replacing this time in the formula of x
[tex]x=v_ocos\theta\frac{2v_osin\theta}{g}[/tex]
We can solve for [tex]v_o[/tex]
[tex]\displaystyle v_0=\sqrt{\frac{xg}{sin2\theta}}[/tex]
Knowing that x=2.20 m and [tex]\theta=54[/tex]°
[tex]\displaystyle v_0=\sqrt{\frac{xg}{sin2\theta}}=4.761m/s[/tex]
We now compute t
[tex]t=\frac{2v_osin\theta}{g}=0.786\ sec[/tex]
