Calculus Help HELP

Question

Consider the following descriptions of the vertical motion of an object subject only the acceleration due to gravity. (Note: the acceleration due to gravity is 9.8 m/s²)

 

a. Find the velocity of the object for all relevant times.

b. Find the position of the object for all relevant times.

c. Find the time when the object reaches its highest point (What is the height?)

d. Find the time when the object strikes the ground.

 

A payload is released at an elevation of 600 m from a hot-air balloon that is rising at a rate of 11 m/s.

 

Answer 1

The initial conditions are y₀ = 600 and v₀ = 11.

 

 

a) a = dv/dt = -g

 

Integrating to get the velocity, we get v = c -gt.

 

At t = 0, v = 11.  So c = v_0.

 

v(t) = v₀ - gt = 11 - 9.8t

 

 

 

b) v = dy/dt = v₀ - gt

 

Integrating, we get y = c + v₀t - ½gt².  At t = 0, y = 600.  So c = y_0.

 

y(t) = y₀ + v₀t - ½gt² = 600 + 11t - 4.9t²

 

 

 

c) The maximum height is reached when v = 0.

 

The time that this occurs satisfies 0 = v₀ - gt.

 

Solving for t, t = v₀/g = 11/9.8 = 1.1 seconds.

 

 

 

 

d)  To find the time, y = 0 when the load hits the ground.

 

0 = y₀ + v₀t - ½gt² = 600 + 11t - 4.9t²

 

Solving for t, t = 12.2 seconds.

Answer 2

a.  v(t) = (-9.8 m/s²)t + 11 m/s

 

b.  h(t) = (1/2)(-9.8)t² + 11t + 600

 

c.  dh(t)/dt = (-9.8)t + 11

0 = -9.8t + 11

9.8t = 11

t = 11/9.8 sec ≅ 1.122 sec

 

To find the height, plug t = 1.122 sec into the h(t) equation

 

d.  h(t) = 0 = (1/2)(-9.8)t² + 11t + 600

 

Use the quadratic formula:

 

t = 11/9.8 ± (1/9.8)√{11² - 4(-4.9)(600)}

t ≅ 1.122 ± (0.010)√11881

t = 1.122 ± 10.9

t = 12.022 and -9.778 sec

 

t = 12.022 sec