Antiderivative/integrals question using velocity

Question

Use a(t) = -32 ft/sec² as the acceleration due to gravity. (Neglect air resistance.)

With what initial velocity must an object be thrown upward (from ground level) to reach the top of a national monument (510 feet)? (Round your answer to three decimal places.)

Answer 1

a(t) = -32 ft/sec²

v(t)=-32t + v₀

s(t) = -16t²+v₀t + s₀

The initial height is 0 since the object is thrown from ground level.

When is the height equal to 0?

-16t²+v₀t = 0

t(-16t + v₀)= 0

t = 0 or t = v₀/16

This means the roots of the downward opening parabola are located at (0,0) and (v₀/16, 0).

The max height will be reached at the axis of symmetry of the parabola. Either find the average of the x-coordinates of the roots or use the formula x = -b/2a to find the x-coordinate of the axis of symmetry. That will be when t = v₀/32.

Plug that value into s(t) to find the value of v₀ when s(v₀/32) = 510.

-16(v₀/32)² + v₀(v₀/32) = 510

This results in something like:

-v₀²/64 + v₀²/32 = 510

Multiply through by 64 to get:

v₀²=(64)(510), then take the square root of both sides (positive square root) to get the value of the initial velocity that will take the object to a height of 510 feet.

desmos.com/calculator/zds0wt2cgc

Answer 2

Given a is constant you can get two equations from taking the antiderivative and using the fundamental theorem of calculus.

Δv = ∫₀^t a dt

(1) v(t) = v_i + a t

Δx = ∫₀^t v(t) dt

(2) x(t) = x₀ + v₀t + 1/2 a t².

(Note Δx = x_f - x_i or if we consider x_f as a function of time, x_f = x(t). Likewise with v.)

If you take we are looking at the specific time when the object reaches the top. (1) and (2) become

(1) v_f = v_i + a t

(2) x_f = x_i + v_it + 1/2 a t².

There is a subtle change in t as well, it is no longer a parameter representing all time, it is the time when the object reaches the top.

If you solve (1) for time and put it into (2) you get

v_f² = v_i² + 2 a (x_f - x_i).

If you've taken physics this is the third equation from kinematics. The minimum speed needed to reach the top gives the condition that the object stops at the top. If you increased the initial speed the object would stop above the desired points. Through specifics of the problem this gives us v_f = 0 ft/s. Plugging in a = -32 ft/s², v_f = 0 ft/s, x_i = 0ft and x_f = 510 ft you should be able to solve for v_i.