A cannonball is launched horizontally off a 20 m high castle wall with a speed of 85 m/s.
How long will the cannonball be in flight before striking the ground?

What is the range of the cannonball?

The ball will be in flight for 2.0 s before it strikes the ground.

The range of the cannonball is 170 m.

Explanation:

Hi there!

The position vector of the cannonball at a time t is given by the following equation:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector of the cannonball at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference on the ground, at the edge of the wall so that the initial position vector is r0 = (0, 20) m.

Using the equation of the vertical component of the position vector r, we can find the time it takes the ball to reach the ground:

y = y0 + 1/2 · g · t²

When the cannonball reaches the ground, y = 0:

0 = 20 m - 1/2 · 9.8 m/s² · t²

-20 m / -4.9 m/s² = t²

t = 2.0 s

The ball will be in flight for 2.0 s before it strikes the ground.

Now, we can calculate the horizontal component of the position vector when the ball reaches the ground at t = 2.0 s (i.e. the range of the cannonball).

x = x0 + v0 · t   (x0 = 0 because we placed the origin of our frame of reference at the wall).

x = v0 · t

x = 85 m/s · 2.0 s

x = 170 m

The range of the cannonball is 170 m.

gravity

explanation:

if they have a finite resultant