A stone is kicked 8m/s horizontally from a cliff 80m high how for from the base of the cliff will the stone strike the ground
Answer (1)
[tex]\textbf{Vertical Motion:} \\[/tex][tex]d = v_0t + \frac{1}{2}at^2 \\[/tex][tex]80 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2 \\[/tex][tex]80 \text{ m} = 4.9 \text{ m/s}^2 \cdot t^2 \\[/tex][tex]t^2 = \frac{80 \text{ m}}{4.9 \text{ m/s}^2} \approx 16.33 \text{ s}^2 \\[/tex][tex]t \approx \sqrt{16.33 \text{ s}^2} \approx 4.04 \text{ s} \\[/tex][tex]\textbf{Horizontal Motion:} \\[/tex][tex]\text{distance} = \text{horizontal velocity} \times \text{time} \\[/tex][tex]\text{distance} = 8 \text{ m/s} \times 4.04 \text{ s} \\[/tex][tex]\text{distance} \approx 32.32 \text{ m} \\[/tex][tex]\textbf{Answer:} \\[/tex][tex]\text{The stone will land approximately 32.32 m from the base of the cliff.}[/tex]
