Question

An aircraft shell is fired vertically upward with a muzzle velocity of 1000ms.calculate;

a=the maximum height it can attain

b=the time taken to reach this height

c=the instantaneous velocities at the ends of 20sec and 50sec

d=when will it's height be 37.5km.interprete your result. neglect air resistance.

Answer 1

(a) Let's take the upwards as the positive direction. Then, we can find the maximum height from the kinematic equation:

$$v^2=v_0^2+2gh,$$ $$0=v_0^2+2gh,$$ $$h=\dfrac{-v_0^2}{2g}=-\dfrac{(1000\ \dfrac{m}{s})^2}{2\times(-9.8\ \dfrac{m}{s^2})}=51\ km.$$

(b) We can find the time that shell takes to reach the maximum height from the kinematic equation:

$$v=v_0+gt,$$ $$0=v_0+gt,$$ $$t=\dfrac{-v_0}{g}=\dfrac{-1000\ \dfrac{m}{s}}{-9.8\ \dfrac{m}{s^2}}=102\ s.$$

(c) We can find the instantaneous velocity at the end of $20\ s$ from the kinematic equation:

$$v=v_0+gt=1000\ \dfrac{m}{s}+(-9.8\ \dfrac{m}{s^2})\times20\ s=804\ \dfrac{m}{s}.$$

We can find the instantaneous velocity at the end of $50\ s$ from the kinematic equation:

$$v=v_0+gt=1000\ \dfrac{m}{s}+(-9.8\ \dfrac{m}{s^2})\times50\ s=510\ \dfrac{m}{s}.$$

(d) We can find the time the shell takes to reach the height of 37.5 km from the kinematic equation:

$$y=v_0t-\dfrac{1}{2}gt^2,$$ $$37500=1000t-4.9t^2,$$ $$4.9t^2-1000t+37500=0.$$

This quadratic equation has two roots: $t_1=154.6\ s$ and $t_2=49.5\ s$. The time $t=49.5\ s$ corresponds to the case where the shell first reaches the height of 37.5 km (as it rises up). The time $t=154.6\ s$ corresponds to the case where the shell secondly reaches the height of 37.5 km (as it falls down).