Question

maximum height reached by a bullet fired vertically upward with a speed equal to 50% if the escape velocity from earth's surface is (R is radius of earth):

Answer 1

Given:

A bullet is fired vertically upward with speed equal to 50% of the escape velocity from earth's surface.

To Find:

Maximum height reached by bullet.

Solution:

Escape Velocity-  Escape velocity is the minimum speed needed for  a non-propelled object to escape from the gravitational influence of a massive body.

Escape velocity of Earth

$V_{e}= \sqrt{\frac{2GM}{R} }$

Where G is gravitational constant , R is radius of earth and M is mass of earth.

Bullet is fired by speed = half of escape velocity

$V = \frac{\sqrt{\frac{2GM}{R} } }{2}$

Maximum Height

Use 3rd equation of motion;

$v^{2} -u^{2} = 2as$

$u=\frac{V_{e} }{2}$

$v = 0 \\a = -g\\s = H_{max}$

Substitute the values;

$(\frac{V_{e} }{2} )^{2} = 2gH_{max}$

$H_{max} = \frac{V_{e} ^{2} }{8g}$

$V_{e}= \sqrt{2gR}$

$H_{max} = \frac{2gR}{8g} \\\\H_{max}= \frac{R}{4}$

Maximum height reached by bullet = $\frac{R}{4}$

Answer 2

The maximum height reached by a bullet fired vertically upward is R/3

Explanation:

The speed of the bullet is:

v = 50% of Ve

The escape velocity is given by the formula:

Ve = 50/100 × √((2GM)/R)

Now, the velocity becomes,

v = 1/2 × √((2GM)/R) → (equation 1)

On applying energy conservation on bullet, we get,

Kinetic energy due to firing = Potential energy experienced by the bullet

- (GMm)/R + 1/2 mv² = - (GMm)/(R + h)

- (GM)/R + 1/2 mv² = - (GM)/(R + h)

1/2 v² = - (GM)/(R + h) + (GM)/R

v² = 2 × (- (GM)/(R + h) + (GM)/R)

On substituting equation (1) in above equation, we get,

(1/2 × √((2GM)/R))² = 2 × (- (GM)/(R + h) + (GM)/R)

1/4 × (2GM)/R) = 2 × (- (GM)/(R + h) + (GM)/R)

1/4 × 1/R = - 1/(R + h) + 1/R

1/4R = h/R(R + h)

R(R + h)/h = 4R

(R + h)/h = 4

R + h = 4h

R = 3h

∴ h = R/3