Answer:
[tex]v = 1.11 \times 10^3 m/s[/tex]
Explanation:
By energy conservation law we will have
[tex]U_i + KE_i = U_f + KE_f[/tex]
as we know that as the projectile is rising up then due to gravitational attraction of earth it will slow down
At the highest position the speed of the projectile will become zero
So here we will have
[tex]-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0[/tex]
[tex]-\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(100)}{6.37 \times 10^6} + \frac{1}{2}(100)v^2 = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{-11})(100)}{(6.37 \times 10^{6} + 65000)}[/tex]
[tex] - 6.25 \times 10^7 + 0.5 v^2 = -6.19 \times 10^7[/tex]
[tex]v = 1.11 \times 10^3 m/s[/tex]
