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  • A ball is dropped from the top of a 100 m high tower on a planet. In the last s before hitting the ground, it covers a distance of 19 m. Acceleration

A ball is dropped from the top of a 100 m high tower on a planet. In the last \frac{1}{2} s before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity ( in ms-2) near the surface on that planet is _______.
Option: 1 8
Option: 2 4
Option: 3 2
Option: 4 6
 

Answers (1)

best_answer

Let the total time taken by the particle is t

So distance covered by a particle in t sec=S_t=ut+\frac{1}{2}gt^2=\frac{1}{2}gt^2

 

Similarly, distance covered by particle in (t-\frac{1}{2}) sec=S_{ t-\frac{1}{2}}=\frac{1}{2}g( t-\frac{1}{2})^2

 

So  distance covered by particle in last (1/2) sec of its journey=S_t-S_{ t-\frac{1}{2}}

 

So 19=S_t-S_{ t-\frac{1}{2}}=\frac{1}{2}gt^2-\frac{1}{2}g( t-\frac{1}{2})^2 =\frac{1}{2}g(t-\frac{1}{4})....(1)

 

and S_t= \frac{1}{2}gt^2=100..(2)

 

from equations (1) and (2) 

 

we get 19t^2-100t+25=0\Rightarrow t=5 , or \ \ t=0.26

putting t=5 we get g = 8 m/s2


 

Posted by

vishal kumar

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