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

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

$T i m e o f d e s c e n t t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 396.9}{9.8}} \Rightarrow t = 9 s e c a n d h o r i z o n t a l d i s t a n c e S = u \times t S = (\frac{720 \times 5}{18}) \times 9 = 1800 m$

Suggest Corrections

Similar questions

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})

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 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/ $s e c^{2}$ )

Join BYJU'S Learning Program

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 (Take g=9.8 m/sec2)

The correct option is D 9 sec and 1800 m Time of descent t=√2hg=√2×396.99.8⇒t=9 secand horizontal distance S=u×tS=(720×518)×9=1800 m

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

The correct option is D 9 sec and 1800 m

$T i m e o f d e s c e n t t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 396.9}{9.8}} \Rightarrow t = 9 s e c a n d h o r i z o n t a l d i s t a n c e S = u \times t S = (\frac{720 \times 5}{18}) \times 9 = 1800 m$