A helicopter flying horizontally with a veloc | vedclass

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Question

(B) Given: Velocity of helicopter $u = 288 \ km/h = 288 	imes \frac{5}{18} \ m/s = 80 \ m/s$. Acceleration due to gravity $g = 10 \ m/s^2$. Let $h$ b

Vedclass · Accepted Answer

$1280$

Vedclass · Answer

(B) Given: Velocity of helicopter $u = 288 \ km/h = 288 	imes \frac{5}{18} \ m/s = 80 \ m/s$. Acceleration due to gravity $g = 10 \ m/s^2$. Let $h$ be the height of the helicopter and $R$ be the horizontal range of the bomb. The time taken to hit the ground is $t = \sqrt{\frac{2h}{g}}$. The horizontal range is $R = u 	imes t = u \sqrt{\frac{2h}{g}}$. The angle $	heta$ made by the line joining the dropping point and the impact point with the horizontal is given by $	an 	heta = \frac{h}{R}$. Given $	heta = 45^{\circ}$, so $	an 45^{\circ} = 1$, which implies $h = R$. Substituting $R = u \sqrt{\frac{2h}{g}}$, we get $h = u \sqrt{\frac{2h}{g}}$. Squaring both sides: $h^2 = u^2 \frac{2h}{g} \implies h = \frac{2u^2}{g}$. Substituting the values: $h = \frac{2 	imes (80)^2}{10} = \frac{2 	imes 6400}{10} = 1280 \ m$.

$P$. Time of flight	$1$. $3$
$Q$. Time when particle is moving at an angle of $53^{\circ}$ with horizontal in upwards direction	$2$. $10$
$R$. Range of the particle	$3$. $50$
$S$. Maximum height reached by the particle	$4$. $60$

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