Question 6 18 pts One projectile is launched from the side of a building which i

Question

Question 6 18 pts One projectile is launched from the side of a building which is 25 meters above the ground. It is launched at 11 m/s at an angle. of 27 degrees above the horizontal and lands on the ground after it travels somee horizontal displacement (this would be how far it lands horizontally from the base of the building.) Another projectile is launched from the same height at the same speed, but at an angle of 27 degrees below the horizontal and lands on the ground after traveling some horizontal displacement. How much further does the second projectile travel, in meters, compared to the first?

Explanation / Answer

for 1st projectile in vertical direction:

vi = 25*sin 27

= 11.35 m/s

a = -9.8 m/s^2

d = -25 m

use:

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

-25 = 11.35*t + 0.5*(-9.8)*t^2

4.9*t^2 - 11.35*t - 25 = 0

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

a = 4.9

b = -11.35

c = -25

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 = 6.188*10^2

roots are :

t = 3.697 and t = -1.38

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

t = 3.697

In horizontal direction:

v = 25*cos 27

= 22.28 m/s

horizontal distance = v*t

= 22.28 m/s * 3.697

= 82.4 m

for 2ns projectile in vertical direction:

vi = -25*sin 27

= -11.35 m/s

a = -9.8 m/s^2

d = -25 m

use:

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

-25 = -11.35*t + 0.5*(-9.8)*t^2

4.9*t^2 + 11.35*t - 25 = 0

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

a = 4.9

b = 11.35

c = -25

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 = 6.188*10^2

roots are :

t = 1.38 and t = -3.697

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

t = 1.38

In horizontal direction:

v = 25*cos 27

= 22.28 m/s

horizontal distance = v*t

= 22.28 m/s * 1.38 s

= 30.75 m

difference = 82.4 m - 30.75 m

= 51.7 m

Answer: 51.7 m

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