Question

A grasshopper jumps at a 58.0°angle, and lands 3.28 m away.What was its initial velocity?(Unit = m/s)

Answer 1

Given data

*The given angle is

`\theta=58.0^0`

*The distance traveled is R = 3.28 m

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} 3.28=(u^2\sin (2*58.0^0))/(9.8) \\ u=5.98\text{ m/s} \end{gathered}`

Thus, the initial velocity is u = 5.98 m/s