Hello, satishinamdar!
I assume that English is not your first language.
The problem is not clearly stated
\(\displaystyle \;\;\)but I will make some guesses . . .
A stone is thrown from a cliff of a mountain at a height of 171 ft. at velocity 17 ft/sec.
What distance is short from a person standing at 228 ft from that person throwing a stone?
From the nature of the problem, I would guess that it
is a projectile example.
I will assume that the initial velocity is
horizontal.
\(\displaystyle \;\;\)That's only way that it makes sense.
I also assume that the second person is standing 228 feet from the
base of the cliff.
Code:
A
**→
| *
|
171 | *
|
|
-+-------*---------------*
B x D C
:- - - - - 228 - - - - -:
The cliff is \(\displaystyle AB\,=\,171\) feet.
The stone is thrown from A with a horizontal velocity of 17 ft/sec.
The second person is at \(\displaystyle C\), 228 feet from \(\displaystyle B\).
The stone strikes the ground at \(\displaystyle D,\;x\) feet from \(\displaystyle B\).
We are asked for the distance \(\displaystyle DC\).
The horizontal position of the stone is: \(\displaystyle \,x\:=\:17t\)
The vertical position of the stone is: \(\displaystyle \,y\:=\:171\,-\,16t^2\)
It strikes the ground when \(\displaystyle y\,=\,0:\;171\,-\,16t^2\:=\:0\;\;\Rightarrow\;\;t\,=\,\frac{3\sqrt{19}}{4}\)
Then point \(\displaystyle D\) is at: \(\displaystyle \,x\:=\:17\left(\frac{3\sqrt{19}}{4}\right)\:\approx\:55.6\) feet.
Therefore, the stone lands \(\displaystyle \,228\,-\,55.6\:=\:172.4\) feet short.
