I need help with math

Question

A ball is thrown up with a initial velocity of 40ft/s

When does the ball reach 24 feet

When does the ball reach 40 feet

What is the greatest height reach by the ball

When does the ball reach the highest point of its path

When does the ball reach the ground

Answer 1

Let

s(t) be the height of the ball after t seconds,

v(t) be the velocity of the ball after t seconds,

a(t) = g = -32 ft/s² be the (constant) gravitational acceleration of the ball.

I am considering upward direction as positive and downward direction

as negative. For that reason the gravitational acceleration is negative.

Let's start counting time when the ball is initially thrown from height 0.

Then we are given

s(0) = 0 ft (1)

v(0) = 40 ft/s (2)

Let's first calculate v(t) and s(t), and those formulas will then

give us all the answers.

Velocity v(t) is the integral of acceleration:

v(t) = ∫a(t)dt = ∫gdt = gt + C

The constant of integration C is set so as to satisfy (2). So

v(t) = gt + 40

Height s(t) is the integral of velocity:

s(t) = ∫v(t)dt = ∫(gt + 40)dt = gt²/2 + 40t + D

The constant of integration D is set so as to satisfy (1). So

s(t) = gt²/2 + 40t

(i) When does the ball reach 24 feet?

That happens at time t satisfying

s(t) = 24

gt²/2 + 40t = 24

-16t² + 40t - 24 = 0

-2t² + 5t - 3 = 0

t = (-5 ± √(5² - 4×(-2)×(-3)))/2×(-2) = (-5 ± 1)/-4

The ball reaches height 24 ft at time 1s on on its way up and at 1.5 s on its way down.

(ii) When does the ball reach 40 feet?

That happens at time t satisfying

s(t) = 40

gt²/2 + 40t = 40

-16t² + 40t - 40 = 0

-2t² + 5t - 5 = 0

t = (-5 ± √(5² - 4×(-2)×(-5)))/2×(-2) = (-5 ± √(-15))/-4

Since the quantity under the square root is negative, the ball never reaches height of 40 ft.

(iv) When does the ball reach the highest point of its path?

That happens half way between the two times obtained in question (i),

i.e. at time 1.25s

(iii) What is the greatest height reach by the ball?

s(1.25) = -32×1.25²/2 + 40×1.25 = 25 ft

(v) When does the ball reach the ground ?

That happens at double the time it takes to reach the highest point,

that is at time 2.5 s.