(B) Given: Velocity of the aeroplane, $u = 60 \ km/h = 60 \times \frac{5}{18} \ m/s = \frac{50}{3} \ m/s$.Height of the aeroplane, $h = 490 \ m$.Accel

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$\left(\frac{500}{3}\right) \ m$

(B) Given: Velocity of the aeroplane, $u = 60 \ km/h = 60 \times \frac{5}{18} \ m/s = \frac{50}{3} \ m/s$.Height of the aeroplane, $h = 490 \ m$.Acceleration due to gravity, $g = 9.8 \ m/s^2$.The time taken by the bomb to reach the ground is given by $h = \frac{1}{2}gt^2$, so $t = \sqrt{\frac{2h}{g}}$.$t = \sqrt{\frac{2 \times 490}{9.8}} = \sqrt{\frac{980}{9.8}} = \sqrt{100} = 10 \ s$.The horizontal distance covered by the bomb in this time is $R = u \times t$.$R = \left(\frac{50}{3}\right) \times 10 = \frac{500}{3} \ m$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

Medium

$A$ bomb is dropped on an enemy post on the ground by an aeroplane flying horizontally with a velocity of $60 \ km/h$ at a height of $490 \ m$. At the time of dropping the bomb, the horizontal distance of the aeroplane from the enemy post so that the bomb hits the target is

TS EAMCET 2023 All TS EAMCET

A
$\left(\frac{400}{3}\right) \ m$
B
$\left(\frac{500}{3}\right) \ m$
C
$\left(\frac{1700}{3}\right) \ m$
D
$498 \ m$

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3-2.Motion in Plane

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Horizontal Projectile Motion

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$A$ bomb is dropped on an enemy post on the ground by an aeroplane flying horizontally with a velocity of $60 \ km/h$ at a height of $490 \ m$. At the time of dropping the bomb, the horizontal distance of the aeroplane from the enemy post so that the bomb hits the target is

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Derive the formula for the range of a projectile $(R)$. Derive the formula for the maximum range of a projectile.

For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then:

$A$ stone is projected at an angle of $30^{\circ}$ to the horizontal. The ratio of the kinetic energy of the stone at the point of projection to its kinetic energy at the highest point of its flight will be:

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