Anti-Aircraft Shell Vertical Ascent: Calculations and Velocities

Question:

An anti-aircraft shell is fired vertically upwards with a muzzle velocity of 294 ms-1.Calculate

• (1) the maximum height reached by it
• (2) the time taken to reach the height
• (3) the velocities at the of 20^th and 40^th second.
• (4) when will its height be 2450m?

Solution:

(1) Anti-aircraft shell-Maximum height

Condition for maximum height(H) is $${\sin^2\theta=1}$$

$${H=\frac{u^2}{2g}=\frac{294^2}{2(9.8)}}$$

$${H=(15)(294)=4410 \ m =4.41 \ Km}$$

(2) Time taken to reach maximum height

Time taken to reach maximum height is $${t=\frac{u}{g}=\frac{294}{9.8}=30 \ s}$$

(3) Velocity at the end of 20^th second

$${v=u-gt}$$

$${v=294-(9.8)(20)}$$

$${v=98 \ \frac{m}{s}}$$

Velocity at the end of 40^th second

$${v=u-gt}$$

$${v=294-(9.8)(40)}$$

$${v=-98 \frac {m}{s}}$$

The negative sign shows that the shell is falling downward.

(4) When will its height be 2450 m ?

$${h=ut-\frac{1}{2}gt^2}$$

$${2450 =294(t)-\frac{1}{2}(9.8)t^2}$$

$${t^2-60t+500}$$

$${t=10 \ sec , t=50 \ sec}$$

t=10 sec (ascending at height 2450 m) and t=50 sec (falling at the other end-same height).