A grasshopper jumps at a 58.0°angle, and lands 3.28 m away.What was its initial velocity?(Unit = m/s)
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Given data
*The given angle is
[tex]\theta=58.0^0[/tex]*The distance traveled is R = 3.28 m
The formula for the horizontal range is given as
[tex]R=\frac{u^2\sin 2\theta}{g}[/tex]*Here g is the acceleration due to gravity
Substitute the values in the above expression as
[tex]\begin{gathered} 3.28=\frac{u^2\sin (2\times58.0^0)}{9.8} \\ u=5.98\text{ m/s} \end{gathered}[/tex]Thus, the initial velocity is u = 5.98 m/s