NawV484757 NawV484757
  • 03-11-2022
  • Physics
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A grasshopper jumps at a 58.0°angle, and lands 3.28 m away.What was its initial velocity?(Unit = m/s)

A grasshopper jumps at a 580angle and lands 328 m awayWhat was its initial velocityUnit ms class=

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AkshaX785588 AkshaX785588
  • 03-11-2022

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

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