Question

A celebrated 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)

Answer 1

`v_0`=4.761 m/s

t=0.786 sec

Step-by-step explanation:

In a projectile motion (or 2D motion), the object is launched with an initial angle
`\theta` and an initial velocity
`v_0`

The components of the velocity are

`v_(ox)=v_ocos\theta`

`v_(oy)=v_osin\theta`

Similarly the displacement has the components

`x=v_(ox).t=v_ocos\theta.t`

`y=v_osin\theta.t-(gt^2)/(2)`

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

`0=v_osin\theta.t-(gt^2)/(2)`

Dividing by t (assuming t different from zero)

`0=v_osin\theta-(gt)/(2)`

Then we find the total flight as

`t=(2v_osin\theta)/(g)`

Replacing this time in the formula of x

`x=v_ocos\theta(2v_osin\theta)/(g)`

We can solve for
`v_o`

`\displaystyle v_0=\sqrt{(xg)/(sin2\theta)}`

Knowing that x=2.20 m and
`\theta=54`°

`\displaystyle v_0=\sqrt{(xg)/(sin2\theta)}=4.761m/s`

We now compute t

`t=(2v_osin\theta)/(g)=0.786\ sec`