Kinematics

JEE Main level, single correct, medium

225 of 308

Question 225 of 308

The trajectory of projectile, projected from the ground is given by y=x−x220y=x-\dfrac{{x}^{2}}{20}. Where xx and yy are measured in meter. The maximum height attained by the projectile will be.

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Answer: (C) 5m5m

The trajectory of a projectile is parabolic in nature. At maximum height, it can be written that
dydx=0...(1)\dfrac{dy}{dx}=0...\left(1\right)
Substitute the expression for trajectory into equation (1) and solve to calculate the horizontal distance, for which the projectile attains the maximum height.
ddx[x−x220]=0⇒1−x10=0⇒x=10m\begin{matrix}\dfrac{d}{dx}\left[x-\dfrac{{x}^{2}}{20}\right] & = & 0 \\ & \Rightarrow & 1-\dfrac{x}{10}=0 \\ & \Rightarrow & x=10m\end{matrix}
Substitute the value of xx in the given expression for the trajectory to obtain the maximum height.
y=(10−10220)m=5m\begin{matrix}y & = & \left(10-\dfrac{{10}^{2}}{20}\right)m \\ & = & 5m\end{matrix}

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