Question

A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 100 m above ground level, and the ball is fired with initial horizontal speed v0. Assume acceleration due to gravity to be g = 9.80 m/s2 .a. Assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. What is the y position of the cannonball at the time tg/2?b. Given that the projectile lands at a distance D = 200m from the cliff. Find the initial speed of the projectile, v0.c. What is the y position of the cannonball when it is at distance D/2 from the hill? If you need to, you can use the trajectory equation for this projectile, which gives y in terms of x directly:

Answer 1

Answer:

a) At tg/2 the cannonball is 25 m below the launching point (75 m above the ground).

b) The initial speed of the projectile is 44.2 m/s.

c) When the x position of the cannonball is 100 m, the y position is 25 m below the launching point (75 m above the ground).  

Explanation:

Hi there!

The equation of the position vector of the cannonball is the following:

r = (x0 + v0 · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector of the ball.

x0 = initial positon of the ball.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the launching point so that x0 and y0 = 0.

When the ball reaches the ground at t = tg, the vertical component of the position vector will be -100 m (see attached figure). Since the cannonball is fired horizontally, it has no initial vertical velocity, then, v0y = 0. Then, the equation of the vertical component of the position vector at t = tg will be:

-100 m = 1/2 · g · tg²

Solving for tg:

2 · (-100 m / g) = tg²

√(-200 m / -9.80 m/s²) = tg

tg = 4.52 s

At t = tg/2

y = 1/2 · g · tg²/4

y = 1/2 · g · (-200 m / 4 · g)

y = -25.0 m

At tg/2 the cannonball is 25 m below the launching point (75 m above the ground).

b) Using the equation of the horizontal component of the position vector, we can calculate the initial velocity:

x = x0 + v0 · t      (x0 = 0)

at t = tg, x = 200 m

200 m = v0 · 4.52 s

200 m / 4.52 s = v0

v0 = 44.2 m/s

The initial speed of the projectile is 44.2 m/s

c) Let´s calculate the time at which the ball is at an horizontal position of x = 100 m.

x = v0 · t

x/v0 = t

100 m / 44.2 m/s = t

t = 2.26 s   (notice that this time is tg/2)

We have already calculated the y position at that time:

y = -25.0 m

When the x position of the cannonball is 100 m, the y position is 25 m below the launching point (75 m above the ground).

Answer 2

Answer: a) When At tg/2 the cannonball is 25 m below that the launching point then (75 m above the ground).

b) After that The initial speed of the projectile is 44.2 m/s.

c) Now When the x position of the cannonball is 100 m so that the y position is 25 m below the launching point that is (75 m above the ground).

Occurs that The equation of the position vector of the cannonball is the following:

• When r = (x0 + v0 · t, y0 + v0y · t + 1/2 · g · t²)
• That is r = position vector of the ball.
• That is x0 = the initial position of the ball.
• That is v0 = initial horizontal velocity.
• t = time.
• y0 = initial vertical position.
• v0y = initial vertical velocity.
• g = acceleration due to gravity.

Now Let´s take the place of the origin of the frame of reference at the launching point so that is x0 and y0 = 0.

• After that When the ball reaches the ground at t = tg, so that the vertical component of the position vector will be -100 m. When the cannonball is fired horizontally, that it has no initial vertical velocity, then that is, v0y = 0. Then, after that the equation of the vertical will be component of the position vector at the t = tg that is:

1. -100 m = 1/2 · g · tg²
2. Solving for tg:
3. 2 · (-100 m / g) = tg²
4. √(-200 m / -9.80 m/s²) = tg
5. tg = 4.52 s
6. At t = tg/2
7. y = 1/2 · g · tg²/4
8. y = 1/2 · g · (-200 m / 4 · g)
9. y = -25.0 m

• When after that At tg/2 the cannonball is 25 m below mention that the launching point (75 m above the ground).

b) Now Using the equation of the horizontal component of the position vector, that we can calculate by the initial velocity:

1. x = x0 + v0 · t (x0 = 0)
2. at t = tg, x = 200 m
3. 200 m = v0 · 4.52 s
4. 200 m / 4.52 s = v0
5. v0 = 44.2 m/s

• Thus that The initial speed of the projectile will be this 44.2 m/s

c) After that Let´s calculate the time at which the ball is at a horizontal position of x = 100 m.

1. x = v0 · t
2. x/v0 = t
3. 100 m / 44.2 m/s = t
4. When t = 2.26 s (notice that this time is tg/2)
5. Then occurs that We have already calculated the y position at that time:
6. y = -25.0 m

• Also Thus that When the x position of the cannonball is 100 m, the y position is 25 m below that the launching point (75 m above the ground).

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