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(C) The initial vertical component of velocity is $u_y = u \sin 	heta = 50 \sin 30^{\circ} = 50 	imes 0.5 = 25 \, ms^{-1}$.Taking the downward direc

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$7$

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(C) The initial vertical component of velocity is $u_y = u \sin 	heta = 50 \sin 30^{\circ} = 50 	imes 0.5 = 25 \, ms^{-1}$.Taking the downward direction as positive,the displacement $s = 70 \, m$,acceleration $a = g = 10 \, ms^{-2}$,and initial velocity $u_y = -25 \, ms^{-1}$ (since it is directed upwards).Using the equation of motion $s = u_y t + \frac{1}{2} a t^2$:$70 = -25t + \frac{1}{2} (10) t^2$$70 = -25t + 5t^2$$5t^2 - 25t - 70 = 0$Dividing by $5$,we get $t^2 - 5t - 14 = 0$.Factoring the quadratic equation: $(t - 7)(t + 2) = 0$.Since time cannot be negative,$t = 7 \, s$.

$A$ ball is projected upwards from the top of a tower with a velocity of $50 \, ms^{-1}$ making an angle of $30^{\circ}$ with the horizontal. The height of the tower is $70 \, m$. After how many seconds from the instant of throwing will the ball reach the ground?

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