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An object is launched at a velocity of 20 m/s in a direction making an angle of 22° upward with the horizontal.What is the maximum height achieved by the projectile?

User WNG
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1 Answer

1 vote

Given data:

Initial velocity,


u=20\text{ m/s}

Angle of projection,


\theta=22\degree

The maximum height achieved by the projectile is given as,


H=(u^2\sin ^2\theta)/(2g)

Here, g is the acceleration due to gravity.

Substituting all known values,


\begin{gathered} H=\frac{(20\text{ m/s})^2*\sin ^2(22\degree)}{2*(9.8\text{ m/s}^2)} \\ \approx2.86\text{ m} \end{gathered}

Therefore, the maximum height achieved by the projectile is 2.86 m.

User Andre Cytryn
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8.8k points
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