A particle is projected at an angle of 30^{ci | vedclass

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(A) The initial vertical component of velocity is $u_y = u \sin 	heta = 60 \sin 30^{\circ} = 60 	imes 0.5 = 30 \; m/s$.The height traversed in the f

Vedclass · Accepted Answer

$5$

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(A) The initial vertical component of velocity is $u_y = u \sin 	heta = 60 \sin 30^{\circ} = 60 	imes 0.5 = 30 \; m/s$.The height traversed in the first second $(t=1)$ is given by $h_0 = u_y t - \frac{1}{2} g t^2 = 30(1) - \frac{1}{2}(10)(1)^2 = 30 - 5 = 25 \; m$.The time taken to reach the maximum height is $T = \frac{u_y}{g} = \frac{30}{10} = 3 \; s$.The height traversed in the last second before reaching the maximum height is the distance covered between $t = 2 \; s$ and $t = 3 \; s$. This is equivalent to the distance covered in the first second of a particle projected vertically downwards from the maximum height with zero initial velocity,which is $h_1 = \frac{1}{2} g t^2 = \frac{1}{2} 	imes 10 	imes (1)^2 = 5 \; m$.Alternatively,using the formula for distance in the $n^{th}$ second: $S_n = u_y + \frac{a}{2}(2n-1)$. For the interval from $t=2$ to $t=3$,$n=3$,$u_y=30$,$a=-10$: $h_1 = 30 + \frac{-10}{2}(2(3)-1) = 30 - 5(5) = 30 - 25 = 5 \; m$.The ratio $h_0 : h_1 = 25 : 5 = 5$.

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