KCET · Physics · Thermodynamics
A stone is thrown vertically at a speed of \( 30 \mathrm{~ms}^{-1} \) making an angle of \( 45^{\circ} \) with the horizontal.
What is the maximum height reached by the stone? Take \( g=10 \mathrm{~ms}^{-2} \).
- A \( 30 \mathrm{~m} \)
- B \( 22.5 \mathrm{~m} \)
- C \( 15 \mathrm{~m} \)
- D \( 10 \mathrm{~m} \)
Answer & Solution
Correct Answer
(B) \( 22.5 \mathrm{~m} \)
Step-by-step Solution
Detailed explanation
Given, speed \( =30 \mathrm{~ms}^{-1} \), angle of projection \( =45^{\circ} ; g=10 \mathrm{~ms}^{-2} \)
Maximum height reached by the stone is
\(H=\frac{v^{2} \sin ^{2} \theta}{2 g}=\frac{(30)^{2}(\sin 45)^{2}}{2 \times 10}=\frac{900 \times \frac{1}{2}}{2 \times 10}=\frac{900 \times 1}{2 \times 2 \times 10}\)
\(\Rightarrow H=\frac{90}{4}=22.5\)
Therefore, maximum height reached by stone is \( 22.5 \mathrm{~m} \).
Maximum height reached by the stone is
\(H=\frac{v^{2} \sin ^{2} \theta}{2 g}=\frac{(30)^{2}(\sin 45)^{2}}{2 \times 10}=\frac{900 \times \frac{1}{2}}{2 \times 10}=\frac{900 \times 1}{2 \times 2 \times 10}\)
\(\Rightarrow H=\frac{90}{4}=22.5\)
Therefore, maximum height reached by stone is \( 22.5 \mathrm{~m} \).
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