1. A cannon in a fort is 18.4 m above the ground, and it fires a cannonball with

Question

1. A cannon in a fort is 18.4 m above the ground, and it fires a cannonball with an initial the fort velocity of 72.4 m/s at a 34 angle above the horizontal. How far from the base of wall does the camonball strike the ground? Hint: you can't use twice the time to reach the peak of flight as the total flight time because the cannon is above ground level. The method to solve this problem is to calculate the time of flight using d at +vt, using the vertical velocity for w, and solve for t with the quadratic equation (or by completing the square). Then solve for the tunce n the usul manner using the horizontal velocty

Explanation / Answer

1)

in vertical direction:

vi = 72.4*sin 34

= 40.5 m/s

a = -9.8 m/s^2

d = -18.4

use:

d = vi*t + 0.5*a*t^2

-18.4 = 40.5*t + 0.5*(-9.8)*t^2

-18.4 = 40.5*t - 4.9*t^2

4.9*t^2 - 40.5*t - 18.4 = 0

This is quadratic equation (at^2+bt+c=0)

a = 4.9

b = -40.5

c = -18.4

Roots can be found by

t = {-b + sqrt(b^2-4*a*c)}/2a

t = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 2.001*10^3

roots are :

t = 8.697 and t = -0.4318

since t can't be negative, the possible value of t is

t = 8.697 s

in horizontal direction:

v = 72.4 * cos 34

=60.02 m/s

range = v*t

=60.02 m/s * 8.697 s

= 522 m

Answer: 522 m

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