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 the tower.
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Answer

A

Explanation

For time calculation, apply
`s_y = u_yt + 1/2 a_yt^2`
In vertical direction.
In all figures, `s=-40m, a_y = -10m//s^2`
In first figures, `u_y = + 20(sqrt2)cos 45^@ = +20m//s`
In second figure ` u_y = 0`
In third figure, `u_y = -20(sqrt2)cos45^@ = -20m//s`
For horizontal distance,
`x = u_xt `
`u_x = 20 m//s ` in all cases
Substituting the proper values and then solving we get,
In first figure
Time `t_1=5.46`s and horizontal distance `x_1 = 109.2 m`
In second figure
Time `t_2 = 2.83` s and the horizontal distance `x_2 = 56.6m`
In third figure
Time `t_3 = 1.46` s and the horizontal distance `x_3 = 29.2m`.