Answer to Question #191143 in Physics for Adrienne Datu

Question

An aircraft shell is fired vertically upward with an initial velocity of 500

m/s. Neglecting friction, compute:

a.) The maximum height it can reach

b.) The instantaneous velocity at the end of 60 sec.

c.) When will its height be at 10 km?

Answer 1

(a)

$$v^2=v_0^2+2gh,$$ $$h=\dfrac{v^2-v_0^2}{2g},$$ $$h=\dfrac{0-(500\ \dfrac{m}{s})^2}{2\cdot(-9.8\ \dfrac{m}{s^2})}=12.75\cdot10^3\ m=12.75\ km.$$

(b) We can find the instantaneous velocity at the end of 60 seconds from the kinematic equation:

$$v=v_0+gt=500\ \dfrac{m}{s}+(-9.8\ \dfrac{m}{s^2})\cdot60\ s=-88\ \dfrac{m}{s}.$$

The sign minus means that the velocity of the shell is directed downward.

(c) We can find the time that the shell takes to reach the height 10 km from the kinematic equation:

$$y=v_0t+\dfrac{1}{2}gt^2,$$ $$10000=500t-4.9t^2,$$ $$4.9t^2-500t+10000=0.$$

This quadratic equation has two roots: $t_1=74.73\ s$ and $t_2=27.31\ s$. The first root corresponds to the case when the shell reaches the height 10 km second time (after it reaches its maximum height and begins to fall down), the second root corresponds to the case when the shell reaches the height 10 km first time (during its path to the maximum height). Therefore, the shell will be at 10 km twice: at $t=27.31\ s$ and at $t=74.73\ s$.