A body is thrown from the surface of the earth with velocity u m s- 1 . The maximum height in meter above the surface of the earth up to which it will reach is ( R= radius of the earth, g= acceleration due to gravity)

Question

A body is thrown from the surface of the earth with velocity u m s- 1 . The maximum height in meter above the surface of the earth up to which it will reach is ( R= radius of the earth, g= acceleration due to gravity) (A) (u2 R/2 g R - u2) (B) (2 u2 R/g R - u2) (C) (u2 R2/2 g R2 - u2) (D) (u2 R/g R - u2)

Tardigrade Admin · Accepted Answer

(- G M m/R) + (1/2) m u2 = 0 + (. - (G M m/R + h) .) (G M/R + h) = (G M/R) - (u2/2) (G M/(. R + h .)) = (2 GM - R u2/2 R) (R + h/G M) = (2 R/2 G M - R u2) h = (2 G M R/2 G M - R u2) - R = (2 G M R - 2 G M R + R2 u2/2 G M - R u2) = (R2 u2/2 G M - R u2) = (R u2/2 g R - u2)

Q. A body is thrown from the surface of the earth with velocity $u m s^{- 1}$ . The maximum height in meter above the surface of the earth up to which it will reach is ( $R =$ radius of the earth, $g =$ acceleration due to gravity)

$\frac{u ^{2} R}{2 g R - u ^{2}}$

$\frac{2 u ^{2} R}{g R - u ^{2}}$

$\frac{u ^{2} R ^{2}}{2 g R ^{2} - u ^{2}}$

$\frac{u ^{2} R}{g R - u ^{2}}$

Q. A body is thrown from the surface of the earth with velocity ums−1 . The maximum height in meter above the surface of the earth up to which it will reach is ( R= radius of the earth, g= acceleration due to gravity)

2gR−u2u2R​

gR−u22u2R​

2gR2−u2u2R2​

gR−u2u2R​

R−GMm​+21​mu2=0+(−R+hGMm​) R+hGM​=RGM​−2u2​ (R+h)GM​=2R2GM−Ru2​ GMR+h​=2GM−Ru22R​ h=2GM−Ru22GMR​−R =2GM−Ru22GMR−2GMR+R2u2​ =2GM−Ru2R2u2​=2gR−u2Ru2​

Q. A body is thrown from the surface of the earth with velocity $u m s^{- 1}$ . The maximum height in meter above the surface of the earth up to which it will reach is ( $R =$ radius of the earth, $g =$ acceleration due to gravity)

$\frac{u ^{2} R}{2 g R - u ^{2}}$

$\frac{2 u ^{2} R}{g R - u ^{2}}$

$\frac{u ^{2} R ^{2}}{2 g R ^{2} - u ^{2}}$

$\frac{u ^{2} R}{g R - u ^{2}}$