Using the equations of kinematics, we obtain the following results;
- Maximum height of the cannonball = 59 m
- Velocity with which the ball strikes the building = 24 m/s
- Total time of flight of the cannonball = 7 secs
<h3>What is the maximum height?</h3>
The maximum height is the height that was attained by the projectile. In this case, we have the fact that the projectile was fired vertically upwards with a speed of 34.0m/s. On the descent, the cannonball lands on top of a 30.0m tall building.
1) The maximum height could be obtained from;
v^2 = u^2 -2gh
v = final velocity ( 0 m/s at the maximum height)
u = initial velocity = 34.0m/s
g = acceleration due to gravity = 9.8 m/s^2
h = maximum height attained
Since v = 0 m/s
u^2 = 2gh
h = u^2/2g
h = (34)^2/ 2 * 9.8
h = 1156/19.6
h = 59 m
2) Given that the distance covered from the maximum height to the top of the building = 59 m - 30 m = 29 m
v^2 = u^2 + 2gh
In this case, the initial velocity is zero because the cannonballs dropped from a height
v^2 = 2gh
v = √2gh
v = √2 * 9.8 * 29
v = 24 m/s
Given that;
v = u - gt
v = 0 m/s when the cannon ball is projected upwards;
u = gt
t = u/g
t = 34.0m/s/9.8 m/s^2
t = 3.5 secs
The total time spent in air = 2(3.5 secs) = 7 secs
Learn more about kinematics:brainly.com/question/7590442
#SPJ1