Explanation:
It is given that,
The horizontal speed of a cliff diver, [tex]v_x=3.6\ m/s[/tex]
It reaches the water below 2.00 s later, t = 2 s
Let [tex]d_x[/tex] is the distance where the diver hit the water. It can be calculated as follows :
[tex]d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m[/tex]
Let [tex]d_y[/tex] is the height of the cliff. It can be calculated using second equation of motion as follows :
[tex]d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m[/tex]
So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.
The height of the cliff is 19.6m and the base of the cliff did the diver hit the water is 7.2m
The distance at the base of the cliff did that the diver hit the water is the horizontal distance. This is expressed as:
[tex]distance=speed \times time[/tex]
[tex]d_x = vt[/tex]
Given the following parameters
Speed v = 3.60m/s
Time t = 2.0s
Substitute into the formula to have:
[tex]d_x = 3.6 \times 2\\d_x = 7.2m[/tex]
The height of the cliff will be the vertical height derived according to the Newton's law of motion:
[tex]h_y=ut+\frac{1}{2}gt^2\\h_y=0(t) + \frac{1}{2}gt^2\\h_y= \frac{1}{2}(9.8)(2)^2\\h_y=4.9(4)\\h_y=19.6m[/tex]
Hence the height of the cliff is 19.6m
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