A ball is rolled off the edge of a horizontal | vedclass

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(A) Given: Horizontal velocity $u = 4\, m/s$,Time $t = 0.4\, s$,Acceleration due to gravity $g = 10\, m/s^2$.Horizontal distance (Range) $R = u 	imes

Vedclass · Accepted Answer

It hits the ground at a horizontal distance $1.6\, m$ from the edge of the table.

Vedclass · Answer

(A) Given: Horizontal velocity $u = 4\, m/s$,Time $t = 0.4\, s$,Acceleration due to gravity $g = 10\, m/s^2$.Horizontal distance (Range) $R = u 	imes t = 4 	imes 0.4 = 1.6\, m$.Vertical height $h = \frac{1}{2}gt^2 = 0.5 	imes 10 	imes (0.4)^2 = 5 	imes 0.16 = 0.8\, m$.Final vertical velocity $v_y = gt = 10 	imes 0.4 = 4\, m/s$.Final speed $v = \sqrt{u^2 + v_y^2} = \sqrt{4^2 + 4^2} = 4\sqrt{2} \approx 5.66\, m/s$.Angle with horizontal $	an 	heta = \frac{v_y}{u} = \frac{4}{4} = 1 \implies 	heta = 45^o$.Comparing with options,option $A$ is correct.

$A$ ball is rolled off the edge of a horizontal table at a speed of $4\, m/s$. It hits the ground after $0.4\, s$. Which statement given below is true?

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