A projectile is thrown with a velocity of 50 | vedclass

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(D) The vertical component of velocity is $u_y = u \sin 	heta = 50 \sin 53^o = 50 	imes (4/5) = 40 \, m/s$.The vertical displacement at any time $t$

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(D) The vertical component of velocity is $u_y = u \sin 	heta = 50 \sin 53^o = 50 	imes (4/5) = 40 \, m/s$.The vertical displacement at any time $t$ is given by $y = u_y t - \frac{1}{2} g t^2$.For a given height $h$,the equation is $\frac{1}{2} g t^2 - u_y t + h = 0$.This is a quadratic equation in $t$,which has two roots $t_1$ and $t_2$ representing the two times the projectile reaches height $h$.The sum of the roots is $t_1 + t_2 = \frac{-(-u_y)}{\frac{1}{2} g} = \frac{2 u_y}{g} = \frac{2 	imes 40}{10} = 8 \, s$.Checking the options:For $A$: $1 + 7 = 8 \, s$.For $B$: $3 + 5 = 8 \, s$.For $C$: $2 + 6 = 8 \, s$.Since all pairs satisfy the condition $t_1 + t_2 = 8 \, s$,the correct option is $D$.

$A$ projectile is thrown with a velocity of $50 \, m/s$ at an angle of $53^o$ with the horizontal. Determine the instants at which the projectile is at the same height.

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