- #1
frosty8688
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1. At t=0, a stone is dropped from the top of a cliff above a lake. Another stone is thrown downward 1.6 s later from the same point with an initial speed of 32 m/s. Both stones hit the water at the same instant. Find the height of the cliff.
2. d_{1}=\frac{1}{2}gt_{1}^{2}; d_{2}=v_{02}t_{2}+\frac{1}{2}gt_{2}^{2};d_{1}=d_{2}; t_{2}=t_{1}- 1.6s
3. \frac{1}{2}gt_{1}^{2}=v_{02}t_{2}+\frac{1}{2}g(t_{1}-1.6s)^{2}. \frac{1}{2}(9.81 m/s^{2})t_{1}^{2}=(32 m/s)(t_{1}-1.6s)+\frac{1}{2}(9.81 m/s^{2})(t_{1}-1.6s)^{2}. Here is where I get stuck in solving the quadratic.
2. d_{1}=\frac{1}{2}gt_{1}^{2}; d_{2}=v_{02}t_{2}+\frac{1}{2}gt_{2}^{2};d_{1}=d_{2}; t_{2}=t_{1}- 1.6s
3. \frac{1}{2}gt_{1}^{2}=v_{02}t_{2}+\frac{1}{2}g(t_{1}-1.6s)^{2}. \frac{1}{2}(9.81 m/s^{2})t_{1}^{2}=(32 m/s)(t_{1}-1.6s)+\frac{1}{2}(9.81 m/s^{2})(t_{1}-1.6s)^{2}. Here is where I get stuck in solving the quadratic.