Question

A student fires a cannonball vertically upwards with a speed of 34.0m/s. On the descent, the cannonball lands on top of a 30.0m tall building. Determine all unknowns and answer the following questions.
What was the cannonball's maximum height (measured from the ground)?

With what speed did the cannonball strike the building?

What was the cannonball's total flight time?

Answer 1

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

What is the maximum height?

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

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