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Question

(A) Let the ball be projected with an initial velocity $u$ at an angle $	heta = 45^{\circ}$.The vertical displacement $y$ at time $t$ is given by $y

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(A) Let the ball be projected with an initial velocity $u$ at an angle $	heta = 45^{\circ}$.The vertical displacement $y$ at time $t$ is given by $y = (u \sin 	heta)t - \frac{1}{2}gt^2$.Since the ball is at the same height $y$ at $t_1 = 2 \ s$ and $t_2 = 8 \ s$, we have:$(u \sin 	heta)t_1 - \frac{1}{2}gt_1^2 = (u \sin 	heta)t_2 - \frac{1}{2}gt_2^2$$u \sin 	heta (t_2 - t_1) = \frac{1}{2}g(t_2^2 - t_1^2)$$u \sin 	heta = \frac{g(t_1 + t_2)}{2} = \frac{10(2 + 8)}{2} = 50 \ m/s$.Since $	heta = 45^{\circ}$, $u \sin 45^{\circ} = u \cos 45^{\circ} = 50 \ m/s$, so $u_x = 50 \ m/s$.The horizontal distance $d$ between the two points is the horizontal displacement between $t_1$ and $t_2$:$d = u_x(t_2 - t_1) = 50 	imes (8 - 2) = 50 	imes 6 = 300 \ m$.

$A$ ball projected at an angle of $45^{\circ}$ with the horizontal crosses two points at equal heights separated by a distance at times $2 \ s$ and $8 \ s$ respectively. The horizontal distance between the two points is (Acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

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