A particle of mass 'm' is projected with a velocity v= kV e ( k

Question

A particle of mass 'm' is projected with a velocity v= kV e ( k <1) from the surface of the earth. ( V e =. escape velocity ) The maximum height above the surface reached by the particle is: (A) R((k/1-k))2 (B) R (( k /1+ k ))2 (C) ( R 2 k /1+ k ) (D) ( Rk 2/1- k 2)

Tardigrade Admin · Accepted Answer

As the particle is projected with a velocity less than escape velocity, it will go to a maximum height and come back.
From conservation of energy

\frac{m g R h}{R + h} = \frac{1}{2} m (k V_{e})^{2}

\frac{2 g R h}{R + h} = k^{2} (2 g R)

\frac{h}{R + h} = k^{2}

\frac{R + h}{h} = \frac{1}{k ^{2}}

\frac{R}{h} + 1 = \frac{1}{k ^{2}}

\frac{R}{h} = \frac{1}{k ^{2}} - 1

h = \frac{R k ^{2}}{1 - k ^{2}}

Q. A particle of mass $^{'} m^{'}$ is projected with a velocity $v = k V_{e} (k < 1)$ from the surface of the earth.
$(V_{e} =$ escape velocity $)$
The maximum height above the surface reached by the particle is:

$R (\frac{k}{1 - k})^{2}$

$R (\frac{k}{1 + k})^{2}$

$\frac{R ^{2} k}{1 + k}$

$\frac{R k ^{2}}{1 - k ^{2}}$

Q. A particle of mass ′m′ is projected with a velocity v=kVe​(k<1) from the surface of the earth. (Ve​= escape velocity ) The maximum height above the surface reached by the particle is:

R(1−kk​)2

R(1+kk​)2

1+kR2k​

1−k2Rk2​

As the particle is projected with a velocity less than escape velocity, it will go to a maximum height and come back. From conservation of energy R+hmgRh​=21​m(kVe​)2 R+h2gRh​=k2(2gR) R+hh​=k2 hR+h​=k21​ hR​+1=k21​ hR​=k21​−1 h=1−k2Rk2​

Q. A particle of mass $^{'} m^{'}$ is projected with a velocity $v = k V_{e} (k < 1)$ from the surface of the earth.
$(V_{e} =$ escape velocity $)$
The maximum height above the surface reached by the particle is:

$R (\frac{k}{1 - k})^{2}$

$R (\frac{k}{1 + k})^{2}$

$\frac{R ^{2} k}{1 + k}$

$\frac{R k ^{2}}{1 - k ^{2}}$