MHT CET · Physics · Motion In One Dimension
A student is throwing balls vertically upwards such that he throws the \(2^{\text {nd }}\) ball when the \(1^{\text {st }}\) ball reaches maximum height. If he throws balls at an interval of 3 second, the maximum height of the balls is \(\left(\mathrm{g}=10 \frac{\mathrm{m}}{\mathrm{s}^2}\right)\)
- A \(45 \mathrm{~m}\)
- B \(35 \mathrm{~m}\)
- C \(25 \mathrm{~m}\)
- D \(30 \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(45 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
At the highest point \(\mathrm{v}=0\)
\(\begin{aligned}& \therefore \mathrm{u}=\mathrm{u}-\mathrm{gt} \\& \therefore \mathrm{u}=\mathrm{gt}=10 \times 3=30 \mathrm{~ms}\end{aligned}\)
Also \(0=\mathrm{u}^2-2 \mathrm{gh}\)
\(\therefore \mathrm{h}=\frac{\mathrm{u}^2}{2 \mathrm{~g}}=\frac{(30)^2}{2 \times 10}=45 \mathrm{~m}\)
\(\begin{aligned}& \therefore \mathrm{u}=\mathrm{u}-\mathrm{gt} \\& \therefore \mathrm{u}=\mathrm{gt}=10 \times 3=30 \mathrm{~ms}\end{aligned}\)
Also \(0=\mathrm{u}^2-2 \mathrm{gh}\)
\(\therefore \mathrm{h}=\frac{\mathrm{u}^2}{2 \mathrm{~g}}=\frac{(30)^2}{2 \times 10}=45 \mathrm{~m}\)
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