Here you go, hope this kinda helps you out :)
Answer:
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Step-by-step explanation:
Let us take the point of projection of the ball as origin of the coordinate system, the upward direction as positive and down direction as negative.
Initial velocity u with which the ball is projected upwards = + 120 ft/s
Uniform acceleration a acting on the ball is to acceleration due to gravity = - 32 ft/s²
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Using the formula:
v² - u² = 2 a h,
where
u = initial velocity of the ball = +120 ft/s
v = final velocity of the ball at the highest point = 0 ft/s
a = uniform acceleration acting on the ball = -32 ft/s²
h = height attained
Substituting the values we get;
0² - 120² = 2 × (- 32) h
=> h = 120²/2 × 32 = 225 feet
The height of the ball from the ground at its highest point = 225 feet + 12 feet = 237 feet.
Let
and
. By the product rule,

By the power rule, we have
and
, but
are functions of
, so we also need to apply the chain rule:


and we have


So we end up with

Replace
to get everything in terms of
:

We can simplify this by factoring:


What are we trying to solve?