The stone reaches the wall at a height of 1.62 m.
The stone lands at a point 24.5 m from the point of projection.
The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.
The horizontal distance x is traveled with a constant velocity u.
[tex]x=ut[/tex]
Calculate the time taken t.
[tex]t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s[/tex]
The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.
[tex]y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m[/tex]
The height h₁ of the stone above the ground when it reaches the wall is given by,
[tex]h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m[/tex]
When the stone reaches the wall, its height from the ground is 1.62m.
The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,
[tex]h=\frac{1}{2} gt_1^2[/tex]
Calculate the time taken by the stone to reach the ground.
[tex]t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s[/tex]
The horizontal distance traveled by the stone is given by,
[tex]R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m[/tex]
The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.
