Answers:
a) 6.566 m/s
b) 0.947 s
Explanation:
The frog's jump can be modeled as projectile motion and the main equations that will be useful in this situation are: Â
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] Â (1) Â
[tex]V^{2}={V_{o}}^{2} - 2gy[/tex] Â (2) Â
Where: Â
[tex]y[/tex] Â is the height of the frog at a given time Â
[tex]y_{o}=0[/tex] Â is the initial height of the frog (assuming it has jumped from ground)
[tex]V_{o}[/tex] Â is the launch speed (initial speed) of the frog
[tex]\theta=45\°[/tex]  is the angle at which the frog jumped
[tex]t[/tex] Â is the time the frog spends on air during the whole parabolic movement
[tex]g=9.8m/s^{2}[/tex] Â is the acceleration due gravity
[tex]V[/tex] Â is the final speed of the frog at a given time
Knowing this, let's begin with the answers:
In this case we will use equation (2) assuming the final velocity of the frog when it hits the ground is zero ([tex]V[/tex]):
[tex]0={V_{o}}^{2} - 2gy[/tex] Â (3) Â
Then we will isolate [tex]V_{o}[/tex]:
[tex]V_{o}=\sqrt{2gy}[/tex] Â (4) Â
[tex]V_{o}=\sqrt{2(9.8m/s^{2})(2.2 m)}[/tex] Â (5) Â
[tex]V_{o}=6.566 m/s[/tex] Â (6) Â This is the frog's launch speed
Since we already have the value of [tex]V_{o}[/tex] we can find the time the frog spends in the air with equation (1), which is fullfilled when [tex]y=0[/tex]:
[tex]0=V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] Â (7) Â
Isolating [tex]t[/tex]:
[tex]t=\frac{2V_{o}sin\theta}{g}[/tex] Â (8) Â
[tex]t=\frac{2(6.566 m/s)sin(45\°)}{9.8m/s^{2}}[/tex]  (9) Â
Finally:
[tex]t=0.947 s[/tex] Â (10) Â This is the time the frog spends in the air
