Question

A ball is thrown across a playing field from a height of h = 5 ft above the ground at an angle of 45° to the horizontal at the speed of 20 ft/s. It can be deduced from physical principles that the path of the ball is modeled by the function

y= -(32/400)x^2+x+5

where x is the distance in feet that the ball has traveled horizontally.

1) Find the maximum height attained by the ball. (Round your answer to three decimal places.)

2) Find the horizontal distance the ball has traveled when it hits the ground. (Round your answer to one decimal place.)

*Don't us a calculator, solve by completing the square*

Answer 1

height, h=5ft

angle of projection, theta=45 degrees

speed, u=20 ft/sec

path function, y=-(32/400)*x^2 + x + 5 ---(1)

dy/dt=-(32/400)*2*x*dx/dt + dx/dt +0

at maximum height, dy/dt=0

0=-(32/400)*2*x*dx/dt + dx/dt ----(2)

here,

dx/dt = ux=u*cos(theta)

ux=20*cos(45)

ux=14.14 ft/sec

from (2) ===>

0=-(32/400)*2*x*dx/dt + dx/dt

0=-(32/400)*2*x*14.14 + 14.14

===> x=6.25 ft

again from (1)

y=-(32/400)*x^2 + x + 5

Hmax=-(32/400)*6.25^2 + 6.25 + 5

Hmax=8.125 ft

b)

from (1),

y=-(32/400)*x^2 + x + 5

0=-(32/400)*x^2 + x + 5

==> x=16.33ft