A particle is projected vertically upward from the surface of the earth with a speed of $\sqrt{\frac{3}{2} g R}$ , $R$ being the radius of the earth and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)

\frac{3 R}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 R

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $3 R$
$\frac{1}{2} m v_{0}^{2} - \frac{G M m}{R}$
$= - \frac{G M m}{R + h}$ and $g = \frac{G M}{R^{2}}$
Therefore $\frac{1}{2} \times \frac{3}{2} \times gR - Rg = - \frac{R^{2} g}{R + h}$
$- \frac{R}{4} = - \frac{R^{2}}{R + h}$
$R + h = 4 R$
We get $h = 3 R$

Suggest Corrections

Similar questions

Q. A particle is projected vertically upward from the surface of the earth with a speed of

\sqrt{\frac{3}{2} g R}

R

being the radius of the earth and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)

Q. Three particles are projected vertically upward from a point on the surface of the earth with velocities

\sqrt{(2 g R / 3)}, \sqrt{(g R)}, \sqrt{(4 g R / 3)}

respectively where R is the radius of the earth and g is the acceleration due to gravity on the surface of the earth. The maximum heights attained are respectively

h_{1} {, h}_{2}, h_{3}

Q. A particle is projected vertically upward from the surface of the earth with a speed of

\sqrt{\frac{3}{2} g R}

R

being the radius of the earth and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)

A particle is projected vertically upwards with a velocity $v = \sqrt{g R}$ , where $R$ denotes the radius of earth and $g$ is acceleration due to gravity of the surface of earth. The maximum height ascended by the particle is

Q. Three particles are projected vertically upward from a point on the surface of the earth with velocities

\sqrt{2 g R / 3}

\sqrt{g R}

\sqrt{4 g R / 3}

respectively where

R

is the radius of the earth and

g

is the acceleration due to gravity on the surface of the earth. The maximum heights attained are respectively

h_{1}, h_{2}, h_{3}

Join BYJU'S Learning Program

A particle is projected vertically upward from the surface of the earth with a speed of √32gR, R being the radius of the earth and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)

The correct option is D 3R12mv20−GMmR =−GMmR+h and g=GMR2 Therefore 12×32×gR - Rg=−R2gR+h −R4=−R2R+h R+h=4R We get h=3R

A particle is projected vertically upward from the surface of the earth with a speed of $\sqrt{\frac{3}{2} g R}$ , $R$ being the radius of the earth and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)

The correct option is D $3 R$
$\frac{1}{2} m v_{0}^{2} - \frac{G M m}{R}$
$= - \frac{G M m}{R + h}$ and $g = \frac{G M}{R^{2}}$
Therefore $\frac{1}{2} \times \frac{3}{2} \times gR - Rg = - \frac{R^{2} g}{R + h}$
$- \frac{R}{4} = - \frac{R^{2}}{R + h}$
$R + h = 4 R$
We get $h = 3 R$