Question

Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings 100 m apart and of same height of 200 m with the same velocity of 25 ms^-1. When and where will the two bullets collides? (Take, g = 10 ms^-2)

Moderate

Solution

Given, distance between the two buildings, d = 100 m
Height of each tower, h = 200 m
Speed of each bullet, v = 25 ms^-1
The situation can be shown as below
where, x be the vertical distance travelled from the top of the building and t be the time at which they collide. As two bullets are fired toward each other, so their relative velocity will be
$v_{rel} = 25 - (- 25) = 50 {ms}^{- 1}$
Then, time, $t = \frac{d}{v_{rel}} = \frac{100}{50} = 2 s$
The distance or height at which they collide is calculated from equation of motion,
$x = u t + \frac{1}{2} a t^{2}$
The bullet is initially at rest, i.e., u = 0 and as it is moving under the effect of gravity a = $-$ g, so
$x = - \frac{1}{2} g t^{2}$
$x = - \frac{1}{2} \times 10 (2)^{2} = - 20 m$
The negative sign shows that the bullets will collide 20 m below the top of tower, i.e. at a height of (200 - 20) = 180 m from the ground after 2 s.

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