From the top of a tower of height 40,m,a ball | vedclass

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(A) The vertical component of initial velocity is $u_y = u \sin 	heta = 20 \sin 30^{\circ} = 20 	imes 0.5 = 10\,m/s$.Using the equation of motion $s

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$2:1$

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(A) The vertical component of initial velocity is $u_y = u \sin 	heta = 20 \sin 30^{\circ} = 20 	imes 0.5 = 10\,m/s$.Using the equation of motion $s_y = u_y t + \frac{1}{2} a_y t^2$ for the total displacement $s_y = -40\,m$:$-40 = 10t - \frac{1}{2} 	imes 10 	imes t^2$$-40 = 10t - 5t^2$Dividing by $-5$,we get $t^2 - 2t - 8 = 0$.Factoring the quadratic equation: $(t - 4)(t + 2) = 0$.Since time cannot be negative,the total time taken to hit the ground is $t = 4\,s$.The time of flight $T$ (time to return to the same elevation) is given by $T = \frac{2 u \sin 	heta}{g} = \frac{2 	imes 20 	imes \sin 30^{\circ}}{10} = \frac{20}{10} = 2\,s$.The ratio of total time to time of flight is $\frac{t}{T} = \frac{4}{2} = 2$,which is $2:1$.

From the top of a tower of height $40\,m$,a ball is projected upwards with a speed of $20\,m/s$ at an angle of elevation of $30^{\circ}$. The ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the same elevation) is (take $g=10\,m/s^2$).

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