The vertical height of the cliff on which the diver drove off is 33.76 m.
The given parameters;
- horizontal velocity of the diver, v = 8 m/s
- horizontal distance traveled by the diver, X = 21
The time of motion of the cliff diver is calculated as follows;
[tex]t = \frac{X}{v} \\\\t = \frac{21}{8} \\\\t = 2.625 \ s[/tex]
The vertical height of the cliff is calculated as follows;
[tex]h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 + 0.5 \times 9.8 \times 2.625^2\\\\h = 33.76 \ m[/tex]
Thus, the vertical height of the cliff on which the diver drove off is 33.76 m.
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