Question

A cannonball is fired at a cliff 1000 m away and strikes it 250 m above the level ground 5 seconds later.

a. What is the angle (in degrees) with which the cannonball was fired, with respect to the level ground? (g = 10 m/s²)

b. What was the approximate initial speed (in m/s) with which the cannonball was fired? (g = 10 m/s²)

c. Which time (in seconds) below is the closest to the time after firing that the cannonball reaches (or would reach) maximum height? (g = 10 m/s²)

Answer 1

Given:

Horizontal distance covered by the cannonball,

Vertical distance covered by the cannonball,

Time of flight,

Let be the initial velocity of the cannonball

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Consider the vertical motion of the cannonball

Use formula

Initially, the cannonball is rising and g is acting downwards, so put a negative sign for g.

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Consider the horizontal motion of the cannonball

There is no acceleration in horizontal direction, so horizontal velocity remains constant.

Horizontal velocity = Horizontal distance/Time of flight

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Launch angle,

(a) ANSWER:

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(b)

Initial launch speed of the cannonball,

(b) ANSWER:

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(c) Time to reach maximum heigth if the cliff was not present.

Consider the vertical motion of the cannonball

Use formula

When the cannonball reaches maximum height, vertical velocity becomes zero.

(c) ANSWER:

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