An aeroplane moving horizontally with a speed of 720 km/h drops a food pocket, while flying at a height of 396.9 m. the time taken by a food pocket to reach the ground and its horizontal range is (Take g = 9.8 m/ $s e c^{2}$ )

3 sec and 2000 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 sec and 500 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 sec and 1500 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9 sec and 1800 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
9 sec and 1800 m

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 396.9}{9.8}} ≃ 9 s e c a n d u = 720 k m / h r = 200 m / s$
$∴ R = u \times t = 200 \times 9 = 1800 m$

Suggest Corrections

Similar questions

An aeroplane moving horizontally with a speed of 720 km/h drops a food pocket, while flying at a height of 396.9 m.the time taken by a food pocket to reach the ground and its horizontal range is (Take g = 9.8 m/ $s e c^{2}$ )

An airplane moving horizontally with a speed of 720 km/h drops a food pocket while flying at a height of 396.9 m. The time taken by a food pocket to reach the ground and its horizontal range is (Take g = 9.8 m/ $s e c^{2}$ )

Q. An aeroplane moving horizontally with a speed of 720 km/h drops a food packet, while flying at a height of 396.9 m. The time taken by a food packet to reach the ground and its horizontal range is

(T a k e g = 9.8 m / s e c^{2})

An aeroplane moving horizontally with a speed of 720 km/h drops a food pocket, while flying at a height of 396.9 m. the time taken by a food pocket to reach the ground and its horizontal range is (Take g = 9.8 m/ $s e c^{2}$ )

Q. An aeroplane moving horizontally with a speed of

720 k m / h

drops a food packet, while flying at a height of

396.9 m

. The time taken by a food packet to reach the ground and its horizontal range is (Take

g = 9.8 m / {sec}^{2})

Join BYJU'S Learning Program

An aeroplane moving horizontally with a speed of 720 km/h drops a food pocket, while flying at a height of 396.9 m. the time taken by a food pocket to reach the ground and its horizontal range is (Take g = 9.8 m/sec2)

The correct option is D 9 sec and 1800 m t=√2hg=√2×396.99.8≃9 sec and u=720 km/hr=200 m/s ∴ R=u×t=200×9=1800 m

An aeroplane moving horizontally with a speed of 720 km/h drops a food pocket, while flying at a height of 396.9 m. the time taken by a food pocket to reach the ground and its horizontal range is (Take g = 9.8 m/ $s e c^{2}$ )

The correct option is D
9 sec and 1800 m

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 396.9}{9.8}} ≃ 9 s e c a n d u = 720 k m / h r = 200 m / s$
$∴ R = u \times t = 200 \times 9 = 1800 m$