A particle is launched at 30^circ to the horizontal with an initial speed of 60,text m/s . Let h_0 be the height covered in the first second and h_1 be the height covered in the last second before reaching maximum height. Find h_0 : h_1. (Take g=10,text m/s ^2)

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EEJ MAIN Physics QUESTION #5511

Question 1 (Enter numeric answer)

A particle is launched at \(30^\circ\) to the horizontal with an initial speed of \(60\,\text{m/s}\). Let \(h_0\) be the height covered in the first second and \(h_1\) be the height covered in the last second before reaching maximum height. Find \(h_0 : h_1\). (Take \(g=10\,\text{m/s}^2\))

Correct Answer:

Correct Answer Explanation

Vertical component of initial velocity: \(u_y=60\sin30°=30\,\text{m/s}\). Height in first second: \(h_0=u_y\cdot1-\frac{1}{2}g(1)^2=30-5=25\,\text{m}\). Time to reach maximum height: \(T=\frac{u_y}{g}=\frac{30}{10}=3\,\text{s}\). In the last second before maximum height, the particle goes from \(t=2\,\text{s}\) to \(t=3\,\text{s}\). Velocity at \(t=2\): \(v=30-10(2)=10\,\text{m/s}\). Height in last second: \(h_1=10(1)-\frac{1}{2}(10)(1)^2=10-5=5\,\text{m}\). Ratio: \(h_0:h_1=25:5=5:1\). So the answer is \(5\).

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