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# An NCC parade is going at a uniform speed of 6km/h through a place under a berry tree on which a bird is sitting at a height of 12.1m. At a particular instant the bird drops a berry. Which cadet give the distance from the tree at the instant) will receive the berry on his uniform?

Last updated date: 13th Jun 2024
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Hint:The equations which established the relation between initial velocity, final velocity, time, acceleration, and displacement are known as the kinematic equation of motion. On releasing the berry from the tree it initially has upward velocity.

Formula used:
Displacement of the berry, ${\rm{S}}\;{\rm{ = }}\;{\rm{ut + }}\dfrac{1}{2}{\rm{a}}{{\rm{t}}^2}$
Here S is displacement, u is initial velocity, a is acceleration due to gravity, t is the time period.

According to given data,
Speed of NCC cadets $= \;6{\rm{km/h = }}\,{\rm{1}}{\rm{.66m/s}}$
Distance of bird from the ground, S $= 12.1{\rm{m}}$
Berry dropped by the bird have initial velocity, $u = \;0$
Acceleration due to gravity $a\; = \;g\; = 9.8m/{s^2}$
Now by using the equation of motion, we can find the time t taken by berry to reach the ground,
Thus we have;
${\rm{S}}\;{\rm{ = }}\;{\rm{ut + }}\dfrac{1}{2}{\rm{a}}{{\rm{t}}^2}$
By putting all the values in this equation we get,
$\Rightarrow \;12.1\; = 0\; + \dfrac{1}{2} \times 9.8{{\rm{t}}^2}$
$\Rightarrow \;{{\rm{t}}^2}\; = \dfrac{{12.1}}{{4.9}}\; = 2.46$
$\Rightarrow \;{\rm{t}}\; = \;1.57s$
Also the distance move by the cadets, ${\rm{d}} = {\rm{v}} \times \;{\rm{t}}$
$\Rightarrow \;{\rm{d = 1}}{\rm{.66}} \times {\rm{1}}{\rm{.57 = 2}}{\rm{.6m}}$
Thus, the cadet who is 2.6m away from the tree will receive the berry on his uniform.

Note:The acceleration due to the gravity of a body is always directed downwards toward the center of the earth, whether a body is projected upwards or downwards. When a body is falling towards the earth, its velocity increases, g is positive When a body is projected upwards, its velocity decreases, g is negative.