A projectile is thrown with speed 40 , m/s at | vedclass

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(B) The vertical displacement of a projectile is given by $y = (u \sin 	heta)t - \frac{1}{2}gt^2$.Since the projectile is at the same height at $t_1

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{1}{\sqrt{3}}ight)$

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(B) The vertical displacement of a projectile is given by $y = (u \sin 	heta)t - \frac{1}{2}gt^2$.Since the projectile is at the same height at $t_1 = 1 \, s$ and $t_2 = 3 \, s$,the sum of these times is equal to the time taken to reach the maximum height multiplied by $2$,which is the total time of flight for that specific height level.Specifically,$t_1 + t_2 = \frac{2u \sin 	heta}{g}$.Given $u = 40 \, m/s$,$g = 10 \, m/s^2$,$t_1 = 1 \, s$,and $t_2 = 3 \, s$:$1 + 3 = \frac{2 	imes 40 	imes \sin 	heta}{10}$.$4 = 8 \sin 	heta$.$\sin 	heta = \frac{4}{8} = \frac{1}{2}$.Therefore,$	heta = 30^{\circ}$.The angle of projection is $	heta = 	an^{-1}(	an 30^{\circ}) = 	an^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

$A$ projectile is thrown with speed $40 \, m/s$ at an angle $\theta$ from the horizontal. It is found that the projectile is at the same height at $1 \, s$ and $3 \, s$. What is the angle of projection?

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