Question

A connon fires a shell with a speed of 84 ms^-1. when the cannon is inclined at 45 degree, the horizontal distance covered is observed as 630 m. what is the percentage decrease in the horizontal distance observed due to air resistance ?​

Answer 1

12.5%

Explanation:

u=84 m/s, ø=45°

Here,

Horizontal distance if there was no air resistance= Range=

(u²sin2ø) / g=(84) ²/9.8 [sin90°=1]

=720m

So,

decrease= (720-630)m =90m

Hence,

%decrease= (90/720)×100%=12.5%

[report this answer if it's misleading. ]

Answer 2

Answer:

Solution:-

v = 34 m/s

R = 630 m

R $= \frac{u {}^{2} \sin2\theta}{9} \\ \\ = \frac{(84) {}^{2} \times \sin(2 \times 45) }{9.81 } \\ \\ = 719.266 \: m$

Actual range observed = 630 m.

% Decrease

$= \frac{89.266}{719.2 \times 10} \\ \\ = 12.41\%$

Since range of projectile in two dimension in the absence in the absence of once air resistance.

Horizontal speed

$= 4 \cos\theta$

Time of flight

$= \frac{24 \sin\theta}{g}$

Thus, range $\frac{4 {}^{2} }{g} \times \sin2\theta$

Percentage decrease in the horizontal distance observed due to air resistance is 12.41% (approx)