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(C) For a projectile,the acceleration is always $g$ acting vertically downwards. At any instant,the tangential acceleration is $a_t = g \sin 	heta$ a

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$1\, s$ and $3\, s$

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(C) For a projectile,the acceleration is always $g$ acting vertically downwards. At any instant,the tangential acceleration is $a_t = g \sin 	heta$ and the normal acceleration is $a_n = g \cos 	heta$,where $	heta$ is the angle the velocity vector makes with the horizontal. Given that the magnitudes of tangential and normal acceleration are equal,we have $a_t = a_n$. Therefore,$g \sin 	heta = g \cos 	heta$,which implies $	an 	heta = 1$. The angle $	heta$ is the angle of the velocity vector with the horizontal,so $	an 	heta = |v_y / v_x| = 1$. This means $|v_y| = |v_x|$. We know $v_x = u_x = 10\, m/s$ and $v_y = u_y - gt = 20 - 10t$. So,$|20 - 10t| = 10$. Case $1$: $20 - 10t = 10 \Rightarrow 10t = 10 \Rightarrow t = 1\, s$. Case $2$: $20 - 10t = -10 \Rightarrow 10t = 30 \Rightarrow t = 3\, s$. Thus,the magnitudes are equal at $t = 1\, s$ and $t = 3\, s$.

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