In the figures shown, three particles are thrown from a tower of height $40 m$ as shown in figure. In each case find the time when the particles strike the ground and the distance of this point from foot of tower.

5.46 s, 2.83 s, 1.46 s, 109.2 m, 56.6 m, 29.2 m

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5.46 s, 1.46 s, 2.83 s, 109.2 m, 56.6 m, 29.2 m

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1.46 s, 109.2 m, 56.6 m, 29.2 m, 5.46 s, 2.83 s

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56.6 m, 29.2 m, 5.46 s, 2.83 s, 1.46 s, 109.2 m

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Solution

The correct option is A $5.46 s, 2.83 s, 1.46 s, 109.2 m, 56.6 m, 29.2 m$
For the case (I),
Horizontal component of velocity= $v_{x} = 20 m / s$
Vertical component of velocity= $v_{y} = 20 m / s$
For the vertical motion,
$s = v_{y} t + \frac{1}{2} a t^{2}$
$⟹ - 40 = 20 t - 5 t^{2}$
$⟹ t = 5.46 s$
Thus the horizontal distance of point from the point of projection = $v_{x} t = 109.2 m$

For the case (II),
Horizontal component of velocity= $v_{x} = 20 m / s$
Vertical component of velocity= $v_{y} = 0 m / s$
For the vertical motion,
$s = v_{y} t + \frac{1}{2} a t^{2}$
$⟹ - 40 = - 5 t^{2}$
$⟹ t = 2.83 s$
Thus the horizontal distance of point from the point of projection = $v_{x} t = 56.6 m$

For the case (III),
Horizontal component of velocity= $v_{x} = 20 m / s$
Vertical component of velocity= $v_{y} = - 20 m / s$
For the vertical motion,
$s = v_{y} t + \frac{1}{2} a t^{2}$
$⟹ - 40 = - 20 t - 5 t^{2}$
$⟹ t = 1.46 s$
Thus the horizontal distance of point from the point of projection = $v_{x} t = 29.2 m$

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