The cannonball in the illustration follows the parabolic trajectory . How far short of the castle does it land?
A) 5 ft
B) 6 ft
C) 3 ft
D) 4 ft
E) 2 ft

Question

The cannonball in the illustration follows the parabolic trajectory  . How far short of the castle does it land? 
A) 5 ft 
B) 6 ft 
C) 3 ft 
D) 4 ft 
E) 2 ft

Quizplus · Accepted Answer

To find the distance short of the castle that the cannonball lands, we need to find the horizontal distance traveled by the cannonball before it lands. We can use the fact that the time taken by the cannonball to reach the highest point of its trajectory (vertex) is half of the total time of flight.

From the illustration, we can see that the cannonball reaches its maximum height at a horizontal distance of 30 ft. Therefore, the horizontal distance from the cannon to the highest point of the trajectory is also 30 ft. We can use this information to find the time taken by the cannonball to reach the vertex:

v = u + at, where v = 0 (at the highest point), a = -g (acceleration due to gravity), u = vertical component of the initial velocity, t = time taken to reach the vertex
0 = vy - gt
t = vy/g

vy is the vertical component of the initial velocity, which is given as 96 ft/s.

t = 96/g = 9.8 s

The total time of flight is twice the time taken to reach the vertex:

total time = 2t = 19.6 s

Now we can use the horizontal component of the initial velocity to find the horizontal distance traveled by the cannonball:

distance = vx × time taken
vx = 75 cos(45) = 53.03 ft/s
distance = 53.03 × 19.6 = 1040.23 ft

The castle is located 1043 ft from the cannon. Therefore, the cannonball falls short of the castle by:

1043 - 1040.23 = 2.77 ft ≈ 3 ft (to the nearest foot)

The Cannonball in the Illustration Follows the Parabolic Trajectory