KCET · Physics · Motion In Two Dimensions
A body is projected vertically upwards. The times corresponding to height while ascending and while descending are \(\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) respectively. Then the velocity of projection is ( is acceleration due to gravity)
- A \(g \sqrt{t_{1} t_{2}}\)
- B \(\frac{g t_{1} t_{2}}{t_{1}+t_{2}}\)
- C \(\frac{g \sqrt{t_{1} t_{2}}}{2}\)
- D \(\frac{g\left(t_{1}+t_{2}\right)}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{g\left(t_{1}+t_{2}\right)}{2}\)
Step-by-step Solution
Detailed explanation
In case of motion under gravity, time taken to go up is equal to the time taken to fall down through the same distance.
Time of descent \(\left(t_{2}\right)=\) time of ascent \(\left(t_{1}\right)=\frac{u}{g}\)
\(\therefore\) Total time of flight \(\mathrm{T}=\mathrm{t}_{1}+\mathrm{t}_{2}=\frac{2 \mathrm{u}}{\mathrm{g}}\)
\(\Rightarrow \quad \mathrm{u}=\frac{g\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)}{2}\)
Time of descent \(\left(t_{2}\right)=\) time of ascent \(\left(t_{1}\right)=\frac{u}{g}\)
\(\therefore\) Total time of flight \(\mathrm{T}=\mathrm{t}_{1}+\mathrm{t}_{2}=\frac{2 \mathrm{u}}{\mathrm{g}}\)
\(\Rightarrow \quad \mathrm{u}=\frac{g\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)}{2}\)
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