Acelebrated mark twain story has motivated contestants in the calaveras county jumping frog jubilee, where frog jumps as long as 2.20 m have been recorded. if a frog jumps 2.20 m and the launch angle is 54.0°, find the frog's launch speed and the time the frog spends in the air. ignore air resistance.
(a) the frog's launch speed (in m/s)
(b)the time the frog spends in the air (in s)

=4.761 m/s

t=0.786 sec

Explanation:

In a projectile motion (or 2D motion), the object is launched with an initial angle and an initial velocity

The components of the velocity are

Similarly the displacement has the components

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

Dividing by t (assuming t different from zero)

Then we find the total flight as

Replacing this time in the formula of x

We can solve for

Knowing that x=2.20 m and °

We now compute t

the answer is a. at the equator

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )