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A particle is dropped from a tower. It's found that it travels \mathrm{45 \mathrm{~m}} in the last second of its journey. Find out the hight of the tower?

\mathrm{(take \, \, g=10 \mathrm{~m} / \mathrm{s}^2 )}

Option: 1

200 m


Option: 2

125 m


Option: 3

370 m


Option: 4

120 m


Answers (1)

best_answer

Let the total lime of journey be n seconds.
using,     \mathrm{ \begin{aligned} S_n=u+\frac{a}{2}(2 n-1) \Rightarrow 45 & =0+\frac{10}{2}(2 n-1) \\ 45 & =5(2 n-1) \\ 45 & =10 n-5 \\ 50 & =10 n \\ n & =5 \end{aligned} }

Height of the tower
                                                                    \mathrm{ \begin{aligned} & h=u t+\frac{1}{2} g t^2 \\ & h=0+\frac{1}{2} \times 10 \times 25 \\ & h=5 \times 25 \\ & h=125 \mathrm{~m} \end{aligned} }
                                                                                                

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