Hi there,
I understand that kinematics can be a challenging topic to grasp at first. Let's break down the problem step by step to help you understand it better.
Firstly, we are given the initial angle of the canon, which is 30 degrees above the horizontal. This means that the ball is being launched at an upward angle, rather than straight ahead. The initial speed of the ball is also given, which is 112.7m/s.
Now, to solve for the maximum height and range, we need to use the equations of kinematics. These equations relate the position, velocity, acceleration, and time of an object in motion.
For the maximum height, we need to use the equation h = u^2sin^2θ/2g, where h is the maximum height, u is the initial speed, θ is the angle of launch, and g is the acceleration due to gravity (9.8m/s^2). Plugging in the values given in the problem, we get:
h = (112.7m/s)^2sin^2(30deg)/2(9.8m/s^2) = 120.3m
Therefore, the maximum height reached by the ball is 120.3m.
To solve for the range, we can use the equation R = u^2sin2θ/g, where R is the range. Plugging in the values, we get:
R = (112.7m/s)^2sin(2(30deg))/9.8m/s^2 = 641.5m
This means that the ball will travel a horizontal distance of 641.5m before hitting the ground.
I hope this explanation helps you understand the problem better. Remember to always write down the given values and use the appropriate equations to solve for the unknowns. If you have any further questions, please don't hesitate to ask. Good luck!
