Question

A stone thrown upward from the top of a 90 ft cliff at 134 ft/sec eventually falls to the beach below?

Answer 1

(a) How long does the stone take to reach its highest point?
time =
(include units)

(b) What is its maximum height?
height =
(include units)

(c) How long before the stone hits the beach?
time =
(include units)

(d) What is the velocity of the stone on impact?
velocity =
(include units)

Answer 2

Gravitational acceleration is \(-32.15 ft/s^2\)

So our height with respect to time is \(x(t) = -32.15t^2 + 134t + 90\)
Our velocity is \(v(t) = x^\prime (t)\)

(a) When we get to the max height, our velocity will be \(0\). So we solve for \(t\) when \(v(t) = x^\prime (t) = 0\).

(b) We substitute the \(t\) we got from (a) into \(x(t)\) to get the height at that time.

(c) We know that when we hit the beach the height was \(0\), so solve tor \(t\) when \( x(t) = 0 \). Use the quadratic equation to solve for \(t\) this time. There will be two solutions. Use the positive one (since the negative time one doesn't make sense).

(d) We substitute the \(t\) we got from (c) into \(v(t) = x^\prime (t)\) to get the velocity at that time.

Answer 3

I leave it to you to find the derivative as well as finding the roots.