Question

an anti-aircraft shell is fired vertically upward with a muzzle velocity of 488 m/s.what is the maximum height it can reach?what time it takes to reach the maximum height? what is the instantaneous velocity at the end of 40 seconds,60 seconds?

Answer 1

Answer on Question 58480, Physics, Other

Question:

An anti-aircraft shell is fired vertically upward with a muzzle velocity of $488\,ms^{-1}$. What is the maximum height it can reach? What time it takes to reach the maximum height? What is the instantaneous velocity at the end of $40\,s$, $60\,s$?

Solution:

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

$$v_f^2 = v_i^2 + 2ah,$$

here, $v_f = 0\,ms^{-1}$ is the final velocity of the shell at the maximum height, $v_i$ is the initial velocity of the shell, $a = g = -9.8\,ms^{-2}$ is the acceleration due to gravity, $h$ is the height.

Then, we get:

$$0 = (488\,ms^{-1})^2 + 2 \cdot (-9.8\,ms^{-2}) \cdot h,$$ $$19.6\,ms^{-2} \cdot h = 238144\,m^2 s^{-2},$$ $$h = \frac{238144\,m^2 s^{-2}}{19.6\,ms^{-2}} = 12.15 \cdot 10^3\,m = 12.15\,km.$$

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

$$v_f = v_i + at,$$

here, $v_f = 0\,ms^{-1}$ is the final velocity of the shell at the maximum height, $v_i$ is the initial velocity of the shell, $a = g = -9.8\,ms^{-2}$ is the acceleration due to gravity, $t$ is the time.

Then, we get:

$$0 = 488\,ms^{-1} + (-9.8\,ms^{-2}) \cdot t,$$ $$9.8\,ms^{-2} \cdot t = 488\,ms^{-1},$$ $$t = \frac {488 \, ms^{-1}}{9.8 \, ms^{-2}} = 49.8 \, s.$$

c) We can find the instantaneous velocity at the end of $40 \, s$ from the kinematic equation:

$$v_f = v_i + at = 488 \, ms^{-1} + (-9.8 \, ms^{-2}) \cdot 40 \, s = 96 \, ms^{-1}.$$

d) Similarly, we can find the instantaneous velocity at the end of $60 \, s$:

$$v_f = v_i + at = 488 \, ms^{-1} + (-9.8 \, ms^{-2}) \cdot 60 \, s = -100 \, ms^{-1}.$$

The sign minus indicates that the velocity of the shell is directed downward (the shell is begin to fall).

**Answer:**

a) $h = 12.15 \, km$

b) $t = 49.8 \, s$

c) $v_f = 96 \, ms^{-1}$

d) $v_f = -100 \, ms^{-1}$

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