A projectile is thrown with velocity u making | vedclass

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Question

(B) Let the vertical component of velocity be $v_y = u \cos 	heta$ (since the angle is with the vertical).The equation for vertical displacement is $

Vedclass · Accepted Answer

$19.6$

Vedclass · Answer

(B) Let the vertical component of velocity be $v_y = u \cos 	heta$ (since the angle is with the vertical).The equation for vertical displacement is $h = v_y t - \frac{1}{2} g t^2$.For the two poles at times $t_1 = 1\,s$ and $t_2 = 3\,s$:$h = v_y(1) - \frac{1}{2} g(1)^2$ ... $(1)$$h = v_y(3) - \frac{1}{2} g(3)^2$ ... $(2)$Equating $(1)$ and $(2)$: $v_y - 0.5g = 3v_y - 4.5g$.$2v_y = 4g \implies v_y = 2g = 2 	imes 9.8 = 19.6\,m/s$.Now,find $h$ using $(1)$: $h = 19.6(1) - 4.9(1)^2 = 14.7\,m$.However,the question asks for the maximum height $H_{max}$ relative to the launch point.$H_{max} = \frac{v_y^2}{2g} = \frac{(19.6)^2}{2 	imes 9.8} = \frac{19.6 	imes 19.6}{19.6} = 19.6\,m$.

$A$ projectile is thrown with velocity $u$ making an angle $\theta$ with the vertical. It just crosses the tops of two poles each of height $h$ after $1\,s$ and $3\,s$,respectively. The maximum height of the projectile is ............ $m$.

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