A bomber dropped a bomb at a height of 490 m. when its velocity along the horizontal was 300 Km h⁻¹.
(a) How long it was in air? 
(b) At what distance from the point vertically below the bomber at the instant the bomb was dropped, did it strike the ground?
 (Ans: 10 s, 833 m)



Given:

Initial velocity along x-axis = `\v_{ix}` = 300 km h⁻¹ = 83.3 m s⁻¹

Initial velocity along y-axis = `\v_{ix}` = 0 m s⁻¹

Acceleration = `\a_x` = 0 m s⁻²

Acceleration = `\a_y` = g = 9.8 m s⁻²

Height = y = h = 10 m


To Find:

(a) Time of flight = t = ?

(b) Horizontal distance covered = x = ?



Solution:

(a) Time of flight = t = ?

To find the time t  here we are using 2nd equation of motion along y-axis

y = `\v_{iy}` t + `\frac {1}{2}` g t²  


by putting the corresponding values

490 m = 0 m s⁻¹ ï½˜ t + `\frac {1}{2}` ｘ 9.8 m s⁻² ｘ t² 

490 m = 0 + 4.9 m s⁻² ｘ t² 

or

t² = `\frac {490 m}{4.9 m s⁻²}`


t² = `\sqrt {100 m s⁻²}`


t = 10 s -------------Ans.(1)

(b) Horizontal distance covered = x = ?

Now we know that along x-axis

x = `\v_{ix}` t

x = 83.3 m s⁻¹ ｘ 10 s

x = 833 m ---------------Ans. (2)


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