# Thread: Determining Rates of Change

1. ## Determining Rates of Change

I need help on how to understand and solve this problem:

A ball is thrown upward at a speed of 48 feet per second from the edge of a cliff 432 feet above ground. Find its height above ground t seconds later.
When does it reach its maximum height?
When does it hit the ground?
(Note: Give the upward direction as positive, physics tells us that the acceleration of the ball is -32 feet per $second^2$.

2. Originally Posted by !!!
I need help on how to understand and solve this problem:

A ball is thrown upward at a speed of 48 feet per second from the edge of a cliff 432 feet above ground. Find its height above ground t seconds later.
When does it reach its maximum height?
When does it hit the ground?
(Note: Give the upward direction as positive, physics tells us that the acceleration of the ball is -32 feet per $second^2$.
Take upwards as positive:

Data:
u = 48 ft/sec
v = 0 ft/sec
a = -32 ft/sec^2
t = ?

Substitute the data into v = u + a t and solve for t.

Data:
u = 48 ft/sec
a = -32ft/sec
x = -432 ft
t = ?

Substitute the data into x = ut + 1/2 a t^2 and solve for t.

3. Originally Posted by !!!
I need help on how to understand and solve this problem:

A ball is thrown upward at a speed of 48 feet per second from the edge of a cliff 432 feet above ground. Find its height above ground t seconds later.
When does it reach its maximum height?
When does it hit the ground?
(Note: Give the upward direction as positive, physics tells us that the acceleration of the ball is -32 feet per $second^2$.
With an accelerated movement you should know that

$v(t) = a\cdot t$ ....... If the acceleration is a then the speed after t seconds is v(t).

$d(t) = \frac12 \cdot a \cdot t^2$ ....... If the acceleration is a then the distance after t seconds is d(t).

$v(0) = 48\ \left[\frac{ft}{s}\right]$

$v(t) = 48 - 32 \cdot t \ \left[\frac{ft}{s}\right]$

$d(t) = 48\cdot t - \frac12 \cdot 32 \cdot t^2 \ [ft] = 48t - 16t^2\ [ft]$ ....... And therefore the height over ground is calculated by:

$h(t) = d(t) + 432\ [ft] = -16t^2 +48t+ 432\ [ft]$

The graph of d(t) is a parabola opening down. So the highest point is reached at the vertex of the parabola which occurs after t = 1.5 s:

$h(1.5) = 468\ [ft]$

The ball hits the ground if the height is zero:

$-16t^2 +48t+ 432 = 0$ ....... Solve for t. You'll get 2 solutions but one of them isn't very plausible. I've got 6.91 s.