A particle is projected at an angle of 30° from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is and height traversed in the last second, before it reaches the maximum height, is . The ratio : is ___________

Question

A particle is projected at an angle of 30° from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is $h_{0}$ and height traversed in the last second, before it reaches the maximum height, is $h_{1}$ . The ratio $h_{0}$ : $h_{1}$ is ___________ [Take, g = 10 $m / s^{2}$ ] [2025]

Smart Achievers · Accepted Answer

(5)

Time to reach maximum height,

$t_{H_{\max}} = \frac{u \sin θ}{g} = \frac{60}{10} \times \frac{1}{2} = 3 s$

Height in $1^{st}$ second $S_{1} = 30 \times 1 - \frac{1}{2} \times 10 \times 1 = 25 m$

Height in last second $S_{3} = 30 + (\frac{- 10}{2}) \times (2 \times 3 - 1) = 5 m$

$\Rightarrow \frac{S_{1}}{S_{3}} = \frac{25}{5} = 5$

Solution

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Solution

Q. A particle is projected at an angle of 30° from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h0 and height traversed in the last second, before it reaches the maximum height, is h1. The ratio h0 : h1 is ___________ [Take, g = 10 m/s2] [2025]

Ans. (5) Time to reach maximum height, tHmax=usinθg=6010×12=3 s Height in 1st second S1=30×1–12×10×1=25 m Height in last second S3=30+(–102)×(2×3–1)=5 m ⇒ S1S3=255=5

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