Question

A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone will reach? Where will the stone land?

Answer 1

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.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

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.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height  h₁ of the stone above the ground when it reaches the wall  is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

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,

$h=\frac{1}{2} gt_1^2$

Calculate the time taken by the stone to reach the ground.

$t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s$

The horizontal distance traveled by the stone is given by,

$R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m$

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.