A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 2.30 seconds after it is fired? (Neglect air resistance.)
Why did the cannon ball go to the doctor?
Because it felt a little projectile!
To find the height of the cannonball above the ground, we can use the kinematic equations of motion.
The given information:
Initial velocity, u = 72.50 m/s (upward)
Time, t = 2.30 seconds
We'll use the following kinematic equation to find the height:
h = u*t + (1/2)*a*t^2
Since the cannonball is shot straight upward, the acceleration, a, will be equal to the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative because it acts in the opposite direction of the initial velocity).
Plugging in the values into the equation:
h = (72.50 m/s) * (2.30 s) + (1/2) * (-9.8 m/s^2) * (2.30 s)^2
Simplifying the equation:
h = 166.375 m + (-25.525 m)
h = 140.85 m
Therefore, the cannonball is approximately 140.85 meters above the ground 2.30 seconds after it is fired.