An underground cannon launches a cannonball from ground level at a 35 degree angle. the cannonball is shot with an initial velocity of 15 m/s.
-how long was the ball in the air for?
-how far did the ball travel?
-what was the max height of the cannonball?

Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]

Explanation:

First, we have to break down the velocity vector into the X & y components.

To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.

In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.

In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation

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