Question #121112
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.
1
Expert's answer
2020-06-17T18:55:00-0400

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


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

acceleration due to gravity is given, g=9.8m/s2g=9.8 m/s^2


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

using equation, v2=u22ghv^2 = u^2 -2gh


0=(20)229.8h0 = (20)^2 - 2*9.8*h

h=20.4mh = 20.4 m above the building.

For height above the ground,

H=20.4+23=43.4mH = 20.4 + 23 = 43.4m


(B) time taken in upward motion, tut_u

v=ugtuv = u - gt_u


putting values,

0=209.8tu    tu=209.80=20-9.8*t_u \implies t_u = \frac{20}{9.8} s


Time taken in downward motion,

using equation, H=12gtd2H = \frac{1}{2} g t_d^2 as initial velocity for downward motion will be zero.

then

td=2Hg=243.49.8t_d = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2*43.4}{9.8}}


total time for motion, t=td+tu=209.8+243.49.8=5.012st= t_d + t_u = \frac{20}{9.8} + \sqrt{\frac{2*43.4}{9.8}}= 5.012 s


2 Given equation is


(3x+2)+(3y3)dydx=0(3x+2) + (3y-3)\frac{dy}{dx} = 0


(3x+2)dx+(3y3)dy=0(3x+2){dx} + (3y-3){dy}= 0

comparing equation with Mdx+Ndy=0Mdx + Ndy = 0


M=3x+2M = 3x+2 and N=3y3N = 3y-3


For equation to be exact, it must follow,


My=Nx\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}


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


Hence equation is exact.


Integrating both sides,


(3x+2)dx+(3y3)dy=C\int (3x+2)dx + \int (3y-3)dy = C where C is constant.


32x2+2x+32y23y=C\frac{3}{2} x^2 + 2x + \frac{3}{2} y^2 - 3y = C



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