King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 48 m/s and at an angle of 29°. A cannonball that was accidentally dropped hits the moat below in 1.7 s.
(a) How far from the castle wall does the cannonball hit the ground?
Perhaps the question setter needs a history lesson? While who was King Arthur is disputed he was certainly around long before (400 years) the first cannon was made in England which was around 1540.
First start with v=u+at to find the height of the cannon.
dude.... that's a slap to the face.... just be quiet
To find the horizontal distance that the cannonball hits the ground, we need to calculate the projectile's horizontal displacement.
First, we need to break the initial velocity into its horizontal and vertical components. The horizontal component (Vx) can be found using the formula Vx = V * cos(θ), where V is the initial velocity of the cannonball (48 m/s) and θ is the angle of projection (29°).
Vx = 48 m/s * cos(29°)
Vx ≈ 41.90 m/s
Now, we need to find the time it takes for the cannonball to hit the ground. Given that the cannonball was accidentally dropped (meaning it falls vertically), its time of flight will be the same as the time it took to hit the moat, which is 1.7 seconds.
Finally, we can use the equation for the horizontal displacement of a projectile to find the distance from the castle wall to where the cannonball hits the ground:
Horizontal displacement (D) = Vx * t
D = 41.90 m/s * 1.7 s
D ≈ 71.023 m
Therefore, the cannonball hits the ground approximately 71.023 meters away from the castle wall.