Solve Quadratic Equation for Cliff Height

[frosty8688]

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.

 

[nil1996]

You should solve the equation yourself as it contains only maths. The equation will give you two values and i think one value of t1 will come negative which you must ignore.Try it :)

 

[frosty8688]

For some reason I get 2.82 seconds for t.

 

[nil1996]

According to my calculation t comes to be 2.02 second

 

[nil1996]

So height of cliff is 20m

 

[nil1996]

Note i have taken g=10m/s^2