A small boy is throwing a ball towards a wall | vedclass

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(A) The motion of the ball can be analyzed by unfolding the path. Since the collisions are perfectly elastic,the trajectory can be treated as a single

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(A) The motion of the ball can be analyzed by unfolding the path. Since the collisions are perfectly elastic,the trajectory can be treated as a single continuous parabola as if the ball were thrown from the ground and reached the same point. Let the total range of this equivalent projectile be $R = 12 \,m$. The equation of the trajectory is $y = x 	an 	heta \left(1 - \frac{x}{R}ight)$.At the release point,$y = 1.4 \,m$ and the horizontal distance from the start is $x$. So,$1.4 = x 	an 	heta \left(1 - \frac{x}{12}ight) \quad \dots(i)$At the wall,$y = 3 \,m$ and the horizontal distance from the start is $6 + x$. So,$3 = (6 + x) 	an 	heta \left(1 - \frac{6 + x}{12}ight) = (6 + x) 	an 	heta \left(\frac{6 - x}{12}ight) \quad \dots(ii)$Dividing $(i)$ by $(ii)$:$\frac{1.4}{3} = \frac{x(12 - x)}{(6 + x)(6 - x)} \Rightarrow \frac{7}{15} = \frac{12x - x^2}{36 - x^2}$$7(36 - x^2) = 15(12x - x^2) \Rightarrow 252 - 7x^2 = 180x - 15x^2$$8x^2 - 180x + 252 = 0 \Rightarrow 2x^2 - 45x + 63 = 0$Solving the quadratic equation: $(x - 21)(2x - 3) = 0$. Since $x < 6$,we get $x = 1.5 \,m$.

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