(A) The initial velocity is given by $\vec{u} = 10 \hat{i} + p \hat{j}$. Thus,$u_x = 10 \ m/s$ and $u_y = p \ m/s$.The maximum height $H$ is given by

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(A) The initial velocity is given by $\vec{u} = 10 \hat{i} + p \hat{j}$. Thus,$u_x = 10 \ m/s$ and $u_y = p \ m/s$.The maximum height $H$ is given by $H = \frac{u_y^2}{2g} = \frac{p^2}{2g}$.The range $R$ is given by $R = \frac{2 u_x u_y}{g} = \frac{2(10)(p)}{g} = \frac{20p}{g}$.According to the problem,$H = 0.5 R$.Substituting the expressions: $\frac{p^2}{2g} = 0.5 \times \frac{20p}{g}$.$\frac{p^2}{2g} = \frac{10p}{g}$.Since $p \neq 0$,we can divide both sides by $p/g$: $\frac{p}{2} = 10$.Therefore,$p = 20$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

Medium

$A$ body is projected into a vertical $X-Y$ plane with the $X$-axis along the horizontal and the $Y$-axis along the vertical with an initial velocity $(10 \hat{i} + p \hat{j}) \ m/s$. If the maximum height reached by the body is $50 \%$ of its range,then the value of $p$ is:

AP EAMCET 2017 All AP EAMCET

A
$20$
B
$10$
C
$5$
D
$2$

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3-2.Motion in Plane

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Horizontal Projectile Motion

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$A$ body is projected into a vertical $X-Y$ plane with the $X$-axis along the horizontal and the $Y$-axis along the vertical with an initial velocity $(10 \hat{i} + p \hat{j}) \ m/s$. If the maximum height reached by the body is $50 \%$ of its range,then the value of $p$ is:

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