Question

A bomb is dropped from a plane flying horizontally with a velocity of 720 kmph at an altitude of 980 m. Time taken by bomb to hit the ground.

Answer 1

Given:

A bomb is dropped from a plane flying horizontally with a velocity of 720 kmph at an altitude of 980 m .

To find:

Time taken by bomb to hit the ground

Solution:

From given, we have,

A bomb is dropped from a plane flying horizontally with a velocity of 720 kmph at an altitude of 980 m .

we use the formula,

h = ut + 1/2 gt²

as initial velocity is zero, we have,

980 = 0 × t + 1/2 × 9.8 × t²

980 = 4.9t²

t² = 980/4.9 = 200

⇒ t = 14.14 sec.

Therefore, the time taken by a bomb to hit the ground is 14.14 sec

Answer 2

Answer

• Time taken = 14.14 s

Given

• A bomb is dropped from a plane flying horizontally with a velocity of 720 km/h at an altitude of 980 m

To Find

• Time taken by bomb to hit the ground

Concept Used

• 2 nd equation of motion

$\boxed{\sf{\green{\bigstar \;\; s=ut+\dfrac{1}{2}at^2}}}$

Solution

Initial velocity , u = 0 m/s

[ Since , bomb dropped from rest ]

Altitude , s = 980 m

Time , t = ? s

Acceleration due to gravity , a = 9.8 m/s²

Apply 2 nd equation of motion .

⇒ s = ut + ¹/₂ at²

⇒ 980 = (0)t + ¹/₂ (9.8)t²

⇒ 980 = 4.9 t²

⇒ t² = 200

⇒ t = 14.14 s

So , Time taken by bomb to hit the ground is 14.14 s

More Info

$\boxed{\begin{minipage}{4.5cm} \bf Equations\ of\ motion\ :\\\\\bf \bigstar \;\; v=u+at \\\\\bf \bigstar \;\; s=ut+\dfrac{1}{2}at^2\\\\\bf \bigstar \;\; v^2-u^2=2as\\\\\bf \bigstar \;\; S_n=u+\dfrac{a}{2}[2n-1] \end{minipage}}$