Question

A dolphin leaps out of the waterat 6.25 m/s at a 45.0° angle.JhingHow far away does it land?(Unit = m)

Answer 1

Given data

*The given angle is

`\theta=45.0^0`

*The given initial velocity is u = 6.25 m/s

The formula for the horizontal range is given as

`R=(u^2\sin 2\theta)/(g)`

*Here g is the acceleration due to gravity

Substitute the values in the above expression as

`\begin{gathered} R=((6.25)^2\sin (2*45.0^0))/(9.8) \\ =3.98\text{ m} \end{gathered}`