Calculus 1 Help

Question

Assume the acceleration of the object is a(t) = −9.8 meters per second. (Neglect air resistance.) With what initial velocity (in m/sec) must an object be thrown upward (from a height of 3 meters) to reach a maximum height of 220 meters? (Round your answer to one decimal place.)

Answer 1

Let

a = -9.8 m/s² be downward acceleration,

t be time variable,

t₀ be the time the object is thrown, which we choose to be 0,

t₁ be the time the object reaches maximum height (unknown),

h(t) be the height of the object as a function of time,

h₀ = h(t₀) = 3m be initial height of the object,

h₁ = h(t₁) = 220m be the final height of the object,

v(t) be the velocity of the object as a function of time,

v₀ = v(t₀) be the sought velocity at h_0,

v₁ = v(t₁) be the velocity at h₁, which is 0,

m is the mass of the object (unknown)

(I) Solution for a physics class:

Use conservation of energy.

Initial energy is the sum of potential energy due to being at height h₀

and kinetic energy due to being thrown with velocity v₀.

Final energy is potential energy due to being at height h₁.

h₀m|a| + mv₀²/2 = h₁m|a|

v₀ = sqrt(2(h₁-h₀)|a|)

(II) Solution for a calculus class:

Acceleration is the derivative of velocity w.r.t. time t, so

velocity v(t) is the integral of (the constant) acceleration a,

so v(t) = at + v₀

Velocity v(t) is the derivative of height h(t) w.r.t time t,

so height h(t) is the integral of v(t), so

h(t) = at²/2 + v₀t + h₀

Expressing both equations at t₁:

v(t₁) = at₁ + v₀

h(t₁) = at₁²/2 + v₀t₁ + h₀

Replacing v(t₁) and h(t₁):

0 = at₁ + v₀

h₁ = at₁²/2 + v₀t₁ + h₀

Substituting t₁ = -v₀/a obtained from the first equation into the second:

h₁ = a v₀²/(2a²) - v₀²/a + h₀

Solving for v₀:

v₀ = sqrt(2(h₀-h₁)a)

Substituting actual numbers:

v₀= sqrt(2(3 - 220)(-9.8)) = 65.2 m/s