A projectile is thrown with a velocity of 10, | vedclass

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(A) Given: Initial velocity $u = 10\,m/s$,angle $	heta = 60^{\circ}$,$g = 10\,m/s^2$.Horizontal component of velocity: $u_x = u \cos 60^{\circ} = 10

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$1$

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(A) Given: Initial velocity $u = 10\,m/s$,angle $	heta = 60^{\circ}$,$g = 10\,m/s^2$.Horizontal component of velocity: $u_x = u \cos 60^{\circ} = 10 	imes 0.5 = 5\,m/s$.Vertical component of velocity: $u_y = u \sin 60^{\circ} = 10 	imes \frac{\sqrt{3}}{2} = 5\sqrt{3}\,m/s$.Let the speed at time $t$ be $v = \sqrt{5g} = \sqrt{50}\,m/s$.The square of the speed is $v^2 = v_x^2 + v_y^2 = 50$.Since $v_x$ remains constant at $5\,m/s$,we have $v_x^2 = 25$.Substituting into the equation: $25 + (u_y - gt)^2 = 50$.$(5\sqrt{3} - 10t)^2 = 25$.Taking the square root: $5\sqrt{3} - 10t = \pm 5$.Case $1$: $10t = 5\sqrt{3} - 5 \implies t_1 = \frac{5(\sqrt{3}-1)}{10} = \frac{\sqrt{3}-1}{2}$.Case $2$: $10t = 5\sqrt{3} + 5 \implies t_2 = \frac{5(\sqrt{3}+1)}{10} = \frac{\sqrt{3}+1}{2}$.The time interval is $\Delta t = t_2 - t_1 = \frac{\sqrt{3}+1 - (\sqrt{3}-1)}{2} = \frac{2}{2} = 1\,s$.

$A$ projectile is thrown with a velocity of $10\,m/s$ at an angle of $60^{\circ}$ with the horizontal. The time interval between the moments when the speed is $\sqrt{5g}\,m/s$ is $..........\,s$ (take $g=10\,m/s^2$).

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