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# The velocity of projection of a projectile is given by : $\vec u = 5\hat i + 10\hat j$ . Find maximum height.(A) $8\,m$(B) $3\,m$(C) $10\,m$(D) $5\,m$

Last updated date: 20th Jun 2024
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Hint From the given equation of the projectile motion, find the horizontal and the vertical component. Use the formula of the maximum height given below, substitute the known values in it, the simplification of it provides the value of the maximum height of the projectile.

Useful formula
The formula for the maximum height of the projectile motion is given by
${H_{\max }} = \dfrac{{{U^2}}}{{2g}}$
Where ${H_{\max }}$ is the maximum height of the projectile motion, $U$ is the vertical component and $g$ is the acceleration due to gravity of the earth.

Complete step by step solution
It is given that the
The velocity of projection of a projectile is $\vec u = 5\hat i + 10\hat j$
In the above equation of the motion, the horizontal component is the $i$ which denotes the distance between the point of the projection and the point of the falling. And the vertical component is the $j$ , which denotes the height of the projectile motion.
Using the formula to find the maximum height,
${H_{\max }} = \dfrac{{{U^2}}}{{2g}}$
Substituting the known values of the vertical component and the acceleration due to gravity as $9.8 \approx 10$ in it,
${H_{\max }} = \dfrac{{{{10}^2}}}{{2 \times 10}}$
By performing the various mathematical operations and simplifying the above equation, we get
${H_{\max }} = \dfrac{{100}}{{20}} = 5\,m$
Hence the value of the maximum height of the projectile motion is obtained as $5\,m$ .

Thus the option (D) is correct.

Note
Let us see the example of the projectile motion. Throwing the ball towards the sky comes under this case. The ball moves up until the force exerted on it there, and when the velocity gets over, the ball comes down by the acceleration due to gravity.