Question

A water balloon is shot into the air with an upward velocity of 95 ft/s. The balloon is released 12 feet above the ground. Write a function that represents the height h of the balloon in feet after t seconds. How long will it take the balloon to reach its maximum height? What will the maximum height be? (ALL ANSWERS SHOULD BE EXPRESSED TO THE NEAREST HUNDREDTH.)

Function: ____________________
Time to reach the maximum height: ___________________
Maximum Height: ______________________
Approximately how many whole seconds will it take to hit the ground?: _________________

Answer 1

The function is: y(t) = 12 + 95*t - 16.085*t²

The time it'll take to reach the maximum height is: 2.95 s

The maximum height is: 152.27 ft

The time it'll take to hit the ground is: 12 s

Explanation:

The water ballon is under the constant acceleration of gravity, therefore it's movement can be modelled by the equations that describe constant acceleration movement, as shown below:

y(t) = y(0) + v(0)*t - 0.5*g*t²

The acceleration is negative, because they are contrary to the movement. Applying the data form the problem, we have:

y(t) = 12 + 95*t - 0.5*32.17*t²

y(t) = 12 + 95*t - 16.085*t²

This equation describes a parabolla, therefore the time at which it achieves the maximum height is the "x" coordinate for the vertex, which can be found by using the formula below:

t = (-b)/[2*a]

t = (-95)/[2*(-16.085)] = -95/(-32.17) = 2.95 s

And the maximum height is:

y(2.95) = 12 + 95*2.95 - 16.085*(2.95)² = 152.27 ft

The time it'll take to reach the ground is:

y(0) = 12 + 95*0 - 16.085*(0)² = 12 s

Answer 2

1. h = h₀ + u×t - 1/2×g×t²

2. 2.95 s

3. 152.14 ft

4. 3.07 s

Explanation:

The given parameters are;

Vertical velocity of the balloon = 95 ft/s

Height from which the balloon is released = 12 feet

The function is h = h₀ + u×t - 1/2×g×t²

Where:

h = Height of the balloon at time t

h₀ = Initial height of the balloon = 12 feet

u = Initial upward velocity of the balloon = 95 ft/s

t = Time duration of motion

g = Acceleration due to gravity = 32.2 ft/s²

2. The time to maximum height is found from;

v = u - g·t

Where:

v = Final velocity = 0 m/s at maximum height

∴ 0 = 95 - 32.2 × t

32.2·t = 95

t = 95/32.2 = 2.95 s

Therefore, the time it will take to reach maximum height = 2.95 s

3. From h = h₀ + u×t - 1/2×g×t², we have;

h = 12 + 95×2.95 - 1/2×32.2×2.95² = 152.14 ft

4. Therefore;

152.14 = 0×t + 1/2 × 32.2 × t²

t² = 152.14/16.1 = 9.45 s

t = √9.45 = 3.07 s