Calculus question dealing with physics

Question

A ball is shot from the ground straight up into the air with initial velocity of  ft/sec. Assuming that the air resistance can be ignored, how high does it go?

Hint: The acceleration due to gravity is 32 ft per second squared.

Answer 1

When it comes to physics problems, what one must do, and a habit which I encourage anyone to develop, is to write-down what one is being given, and what one is being asked to find; therefore:

Given:

v₀ = 45 ft./s.

a = g = 32 ft./s.

Find:

h = ?? ft.

The standard equation which must be used, here, is:

y - y₀ = v₀·t + (1/2) a·t²

Where the values are:

y = final height

y₀ = initial height

v₀ = initial velocity

a = acceleration

t = time

In this case, however, one is being asked to find the total height travelled by the ball, after being shot by the cannon. The total height is h = y - y₀, and the acceleration is that due to gravity a = g; therefore, the kinematic equation becomes:

h(t) = v₀·t + (1/2) g·t²

Substituting the values given into this last equation:

h(t) = 45t + (1/2)(-32)t²; where a = g = -32 ft./s. is negative because it is slowing-down the ball

h(t) = 45t - 16t²

The problem does not state for how long the ball was in flight; therefore, this last equation may not be used, as it currently is, to find the maximum height reached by the ball. However, a known little fact about things which are shot, or thrown, straight-up into the air is that objects stop moving upwards once they no longer have a velocity. No velocity, no movement. So, by taking the derivative of this last equation, one obtains the velocity function:

1. h(t) = 45t - 16t²
2. v(t) = h'(t) = 45 - 32t

This last equation, as it was above-stated, may be used to determine the height travelled by the ball by first determining the time in which the ball was in flight. Since the ball reaches its maximum height when it stops moving (when the velocity is zero), this last equation must be set to zero to find-out the the amount of time in which the ball was in-flight:

1. 45 - 32t = v(t)
2. 45 - 32t = 0
3. -32t = -45
4. t ≈ 1.40625 ft./s.
5. t ≈ 1.4 (two significant figures being used)

This last value of t ≈ 1.4 ft./s. may now be used to determine the height reached by the ball using the earlier obtained equation:

1. h(t) = 45t - 16t²
2. h(t) = (45)(1.4) - (16)(1.4)²
3. h(t) = 63 - 31
4. h(t) ≈ 32 ft. (again, using two significant figures)

FINAL ANSWER:

h = 32 ft.

Answer 2

the height equation is

h(t) = -16t^2 + 45t

take the derivative and set it equal to zero

h'(t) = -32t +45 = 0

solve for t

32t=45 = about 1.4 seconds (about sqr2). It hits the ground in about twice that or 2.8 seconds

t = 45/32 seconds to reach maximum height

plug that value into the original height equation h(t)

h(45/32) = max height = -16(45/32)^2 + 45(45/32) = (45/32)[-16(45)/32 + 45] =

(45/32)[-45/2 +45] = (45/32)(45/2) = 2025/64 = 31.64 feet