Question

A shell is fired horizontally from a gun situated at a height of 44.1m above a horizontal plane with a muzzle
velocity of 300m/s to hit a target on the horizontal plane.(i)How long does the shell take to hit the target ?(ii)
Where is the target situated? (iii)With what vertical velocity does the shell strike the target? g=9.8m/s2

Answer 1

Given:

A shell is fired horizontally from a gun situated at a height of 44.1m above a horizontal plane with a muzzle velocity of 300m/s to hit a target on the horizontal plane.

To find:

• Time taken to hit the target ?
• Location of target ?
• Vertical velocity of striking ?

Calculation:

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$\therefore \: h = u_{y}(t) + \dfrac{1}{2} g {t}^{2}$

$\implies \: 44.1= 0(t) + \dfrac{1}{2} g {t}^{2}$

$\implies \: 44.1= \dfrac{1}{2} g {t}^{2}$

$\implies \: {t}^{2} = \dfrac{44.1 \times 2}{9.8}$

$\implies \: {t}^{2} = 9$

$\implies \: t = 3 \: sec$

So, it will take 3 seconds to hit the target.

Now, range :

$\therefore \: R = u_{x}(t)$

$\implies \: R = 300 \times 3$

$\implies \: R = 900 \: m$

So, target is situated 900 metres from the base of the building.

Now vertical velocity :

$\therefore \: v_{y} = u_{y} + gt$

$\implies\: v_{y} = 0 + g(3)$

$\implies\: v_{y} = 9.8 \times 3$

$\implies\: v_{y} = 29.4 \: m {s}^{ - 1}$

So, vertical velocity of striking is 29.4 m/s.