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