A cannonball is fired horizontally from a 240 {m}-high cliff toward a vertical building 720 {m} away with an initial velocity of 120 {m/s}.

When the ball will hit the building?
in
a. 7.00 {s}
b. 6.00 {s}
c. less than 6.00 {s}
d. more than 7.00 {s}

The ball would take sqrt(2H/g) = 7.00 s to hit the ground, but will reach the building in 720/120 = 6.00 s

So it will not hit the ground first, and will arrive in exactly 6.00 s.

y=h=120m t=d/v y=1/2 g t^2 t=√(240/9.80s^2 ) t=440m/(110 m/s) t^2=2y/g

x=d=440m t=4s t=√(2y/g)
V_0=110 m/s 4s. to hit the building t=√((2(120m))/(9.80 m/s^2 ))
t=√(24.5s^2 )
t= ? To hit the ground in t=4.95s

To determine when the cannonball will hit the building, we can use the formula:

\[d = v_0t + \dfrac{1}{2}at^2\]

Where:
- \(d\) is the horizontal distance traveled by the cannonball (720 m)
- \(v_0\) is the initial horizontal velocity of the cannonball (120 m/s)
- \(t\) is the time elapsed
- \(a\) is the horizontal acceleration (which is 0 since the cannonball is fired horizontally)

Plugging in the provided values, we get:

\[720 = 120t + \dfrac{1}{2}(0)t^2\]

Simplifying the equation, we have:

\[720 = 120t\]

Now, we can solve for \(t\):

\[t = \dfrac{720}{120} = 6\]

Therefore, the cannonball will hit the building after 6 seconds.

The correct answer is b. 6.00 s.

To determine when the cannonball will hit the building, we need to find the time it takes for the cannonball to travel horizontally a distance of 720 m.

First, let's consider the horizontal motion of the cannonball. Since there is no acceleration in the horizontal direction (assuming no air resistance), the horizontal velocity remains constant throughout the motion.

We are given the initial horizontal velocity of the cannonball as 120 m/s. Since there is no acceleration, the vertical velocity and position of the ball do not affect the horizontal motion.

The horizontal distance traveled by an object can be found using the formula:

distance = velocity × time

In this case, the distance is 720 m and the horizontal velocity is 120 m/s. Rearranging the formula, we have:

time = distance / velocity

Substituting the given values, we get:

time = 720 m / 120 m/s = 6 s

Therefore, the cannonball will hit the building in 6.00 s, which corresponds to option (b).