Final answer:
The cannonball, launched horizontally from a 75 m high castle wall at 60 m/s, will be in flight for approximately 3.91 seconds before striking the ground. This is calculated using the kinematic equation for vertical displacement under uniform acceleration due to gravity.
Step-by-step explanation:
The question deals with the time of flight for a cannonball launched horizontally from a certain height. Using the physics of projectile motion, we can determine how long the cannonball will be in the air before it strikes the ground. Based on the information provided, the height from which the cannonball is launched is 75 meters, and it has an initial horizontal velocity of 60 m/s.
Since the cannonball is launched horizontally, its initial vertical velocity is 0 m/s. The only acceleration acting on the cannonball in the vertical direction is the acceleration due to gravity, which is 9.81 m/s2 directed downwards. We use the kinematic equation for uniformly accelerated motion to find the time t:
s = ut + (1/2)at
2
Where:
s = vertical displacement (75 m)
u = initial vertical velocity (0 m/s)
a = acceleration due to gravity (9.81 m/s2)
t = time in seconds.
By substituting the known values, we have:
75 = 0 * t + (1/2) * 9.81 * t2
Solving for t, we find:
t = sqrt((2 * 75) / 9.81)
t ≈ sqrt(15.29)
t ≈ 3.91 seconds
Therefore, the cannonball will be in flight for approximately 3.91 seconds before striking the ground.