Question

a cannon ball fire a shell with a peed of 84m/s.when the canon is incline at 45 degree ,the horizontal distance covered is is observed as 630 m.what is the percentage decrease in horizontal distance observed due to air resistance?

Answer 1

A cannon ball fires a shell with a speed of $84\mathrm{m / s}$ . When the canon is inclining at 45 degree, the horizontal distance covered is observed as $630\mathrm{m}$ . What is the percentage decrease in horizontal distance observed due to air resistance?

Solution.

$v_{x} = v_{0}\cos \alpha ;$

$v_{y} = v_{0}\sin \alpha .$

The max distance.

$l_{max} = v_{0x}t;$

$l_{max} = v_{0}\cos \alpha t;$

The time of flight.

$y = v_{0y}t - \frac{gt^2}{2};$

$y = v_{0}\sin \alpha t - \frac{gt^{2}}{2}.$

At the end of the flight $y = 0$ :

$0 = v_{0}\sin \alpha t - \frac{gt^{2}}{2};$

$v_{0}\sin \alpha t = \frac{gt^2}{2};$

$v_{0}\sin \alpha = \frac{gt}{2};$

$$t = \frac {2 v _ {0} \sin \alpha}{g}.$$

The max distance.

$$l _ {m a x} = v _ {0} \cos \alpha \frac {2 v _ {0} \sin \alpha}{g};$$ $$l _ {m a x} = \frac {2 v _ {0} ^ {2} \sin \alpha \cos \alpha}{g}.$$ $$l _ {m a x} = \frac {2 \cdot 8 4 ^ {2} \cdot \sin 4 5 {}^ {\circ} \cdot \cos 4 5 {}^ {\circ}}{9 . 8} = 7 2 0 (m).$$

The percentage decrease in horizontal distance observed due to air resistance:

$$k = \frac {l _ {\max} - l}{l _ {\max}} \cdot 100 \% = \left(1 - \frac {l}{l _ {\max}}\right) \cdot 100 \% ;$$ $$k = \left(1 - \frac {l}{l _ {\max}}\right) \cdot 100 \%.$$ $$k = \left(1 - \frac {630}{720}\right) \cdot 100 \% = 12.5 (\%).$$

**Answer:**

The percentage decrease in horizontal distance observed due to air resistance is $k = 12.5\%$.