AP EAMCET · PHYSICS · Motion In One Dimension
If a ball projected vertically upwards with certain intitial velocity from the ground crosses a point at a height of 25 m twice in a time interval of 4 s, then the initial velocity of the ball is
(Acceleration due to gravity \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) )
- A \(20 \mathrm{~m} \mathrm{~s}^{-1}\)
- B \(30 \mathrm{~m} \mathrm{~s}^{-1}\)
- C \(40 \mathrm{~m} \mathrm{~s}^{-1}\)
- D \(25 \mathrm{~m} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(B) \(30 \mathrm{~m} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
\(h = ut - \frac{1}{2}gt^2 \Rightarrow 25 = ut - \frac{1}{2}(10)t^2\) \(5t^2 - ut + 25 = 0\) \(t_1 + t_2 = \frac{u}{5}\), \(t_1 t_2 = \frac{25}{5} = 5\) \((t_2 + t_1)^2 = (t_2 - t_1)^2 + 4t_1 t_2\) \((\frac{u}{5})^2 = (4)^2 + 4(5)\) \(\frac{u^2}{25} = 16 + 20 = 36\)…
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