Question

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Answer 1

From the peak to the bottom, using
\[s=ut+\frac{1}{2}at^2\]
we get
\[d+h=vt+\frac{1}{2}at^2\]
\[d+h=\frac{1}{2}(-9.81)(16)^2\]
\[|d+h|=1244.7m\]

Answer 2

I just worken on this problem and found 1000 m ???

Answer 3

There was a mistake there the |d+h| should be 1255.7m
We can use \[v^2=u^2+2ad \]
to find 'd'
So,
\[0=15^2+2(-9.81)(d)\]
\[d=11.5m\]
Then minus it out to get 'h'
\[d+h=1255.7m \\ h=1244.7m\]