Question

A projectile of mass m is fired from the surface of the earth at an angle alpha = 60^(@) from the vertical. The initial speed upsilon_(0) is equal to sqrt((GM_(e))/(R_(e)). How high does the projectile rise ? Neglect air resistance and the earth's rotation.

Answer 1

Rmax(max height)= 3R/2

At max height Rmax let velocity :Vo

Torque will be zero at max height because angle between position vector and force vector is 180 degree.

So angular momentum is conserved

mV0rsinalpha=mVpRmax....(1)

Also mechanical energy is conserved

Hence pe1+ke1 :pe2 +ke2......(2)

From 1&2

Rmax=3r/2 , where r : radius of earth.

Answer 2

Height from earth surface = $\frac{r_e}{2}$

Given:

A projectile of mass m is fired from the surface of the earth at an angle alpha = 60° from the vertical.

Initial speed is equal to$\sqrt{\frac{GM_e}{R_e} }$

To find:

How high does the projectile rise.

Concept used:

Conservation of angular momentum and mechanical energy.

Explanation:

Let $V_0$  be the initial speed of the projectile at highest point and $t_{max}$ its distance from the center if the earth.

Applying conservation of angular momentum from initial position to Maximum height.

$m v_0 r sin \alpha$ = $m v_0 r_{max}$ .......1

Conservation of mechanical energy.

$PE_i + KE_i = PE_f + KE_f$

$\frac{1}{2} m v_{0}^{2}-\frac{G M_{e} m}{r_{e}}=\frac{1}{2} m v^{2}-\frac{G M_{e} m}{r_{\text {max }}}$    .........2

By solving equation 1 and 2 we can get

$r_{max}$ = $\frac{3r_e}{2}$

Height from earth surface =  $r_{max}$ -$r_{e$  = $\frac{r_e}{2}$

To learn more...

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