A stone was thrown from the top of a cliff 60 metres above sea level. The height H metres, of the stone above sea level t seconds after it was released is given by H(t)=-5t^2+20t+60.

a) Find the time given for the stone to reach its maximum height.
b) What is the maximum height above sea level reached by the stone?
c) How long before the stone strikes the water?

Please help!

A problem of this nature would typically be tackled with one of two different approaches. From the title of your post I'll guess that you're to solve this one algebraically, rather than via calculus.

Start by putting your height function H(t) into "vertex form" using the method of "completing the square". You'll end up with the function in the alternate form

H(t) = A(t - B)^2 + C

... where A, B, and C will be actual numbers, and t will remain as a variable.

A will be a negative value in this case, telling you that this quadratic is concave (opens down) and hence its vertex is its unique maximum point, rather than its minimum point. Therefore, the stone's maximum height corresponds to the quadratic's vertex.

More specifically, the value of B gives you the time-coordinate of the vertex, and the value of C gives you the height-coordinate. Hence, (a) and (b) are solved.

To solve (c) you'll need to find the value of t such that height H = 0. Using the vertex form of the formula above, solve for t in

0 = A(t - B)^2 + C

Hint: You'll end up with two solutions, involving the +/- options of an expression involving a square root. Take the positive one, as "negative time" doesn't really have any meaning in this scenario.