Answer to Question #121112 in Differential Equations for Joseph Se

Question

1. A ball with mass m kg is thrown upward with initial velocity 20 m/s from the roof of a building 23 m high. Neglect air resistance.

Use g = 9.8 m/s^2. Round your answers to one decimal place.

(a) Find the maximum height above the ground that the ball reaches.
x max = _______ meters (absolute tolerance +/- 0.1)

(b) Assuming that the ball misses the building on the way down, find the time that it hits the ground.

t tend =______ s (absolute tolerance +/- 0.1)

2. Determine whether the equation is exact. If it is exact, find the solution. If it is not exact, enter NS.

(3x+2)+(3y-3) y prime = 0
Do not enter an arbitary constant.
_________ = c, where c is an arbitrary constant.

Answer 1

1(A) Height of the building is given 23m

Velocity with which ball is projected upward, $u = 20m/s$

acceleration due to gravity is given, $g=9.8 m/s^2$

At maximum height, velocity of ball will be zero. i.e. $v=0$

using equation, $v^2 = u^2 -2gh$

$0 = (20)^2 - 2*9.8*h$

$h = 20.4 m$ above the building.

For height above the ground,

$H = 20.4 + 23 = 43.4m$

(B) time taken in upward motion, $t_u$

$v = u - gt_u$

putting values,

$0=20-9.8*t_u \implies t_u = \frac{20}{9.8}$ s

Time taken in downward motion,

using equation, $H = \frac{1}{2} g t_d^2$ as initial velocity for downward motion will be zero.

then

$t_d = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2*43.4}{9.8}}$

total time for motion, $t= t_d + t_u = \frac{20}{9.8} + \sqrt{\frac{2*43.4}{9.8}}= 5.012 s$

2 Given equation is

$(3x+2) + (3y-3)\frac{dy}{dx} = 0$

$(3x+2){dx} + (3y-3){dy}= 0$

comparing equation with $Mdx + Ndy = 0$

$M = 3x+2$ and $N = 3y-3$

For equation to be exact, it must follow,

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

$\frac{\partial M}{\partial y} = 0$ and $\frac{\partial N}{\partial x} = 0$

Hence equation is exact.

Integrating both sides,

$\int (3x+2)dx + \int (3y-3)dy = C$ where C is constant.

$\frac{3}{2} x^2 + 2x + \frac{3}{2} y^2 - 3y = C$