A cannon is fired with muzzle velocity of 150 m/s at an angle of elevation = 45°. Gravity = 9.8 m/s2.
a) What is the maximum height the projectile reaches?
b) What is the total time aloft?
c) How far away did the projectile land? (Range)
d) Where is the projectile at 10 seconds after firing?

A.) The formulas we will be using are: d = v0t + ½at2 and vf – v0 = at In order to find the distance h, we need to know two things: the velocity at h and the amount of time it takes to get there. The first is easy. The vertical component of the velocity is equal to zero at point h. This is the point where the upward motion is stopped and the projectile begins to fall back to Earth. The initial vertical velocity is v0y = v0·sin v0y = 150 m/s · sin(45°) v0y = 106.1 m/s Now we know the beginning and final velocity. The next thing we need is the acceleration. The only force acting on the

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Question|Asked by Taurus28

Asked 14 days ago|2/9/2024 12:12:51 PM

Updated 14 days ago|2/9/2024 12:49:42 PM

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A.) The formulas we will be using are:

d = v0t + ½at2

and

vf – v0 = at

In order to find the distance h, we need to know two things: the velocity at h and the amount of time it takes to get there. The first is easy. The vertical component of the velocity is equal to zero at point h. This is the point where the upward motion is stopped and the projectile begins to fall back to Earth.

The initial vertical velocity is

v0y = v0·sin

v0y = 150 m/s · sin(45°)

v0y = 106.1 m/s

Now we know the beginning and final velocity. The next thing we need is the acceleration.

The only force acting on the projectile is the force of gravity. Gravity has a magnitude of g and a direction in the negative y direction.

F = ma = -mg

solve for a

a = -g

Now we have enough information to find the time. We know the initial vertical velocity (V0y) and the final vertical velocity at h (vhy = 0)

vhy – v0y = at

0 – v0y = -9.8 m/s2·t

0 – 106.1 m/s = -9.8 m/s2·t

Solve for t

projectile motion math step 3

t = 10.8 s

Now solve the first equation for h

h = v0yt + ½at2

h = (106.1 m/s)(10.8 s) + ½(-9.8 m/s2)(10.8 s)2

h = 1145.9 m – 571.5 m

h = 574.4 m

The highest height the projectile reaches is 574.4 meters.

d = v0t + ½at2

and

vf – v0 = at

In order to find the distance h, we need to know two things: the velocity at h and the amount of time it takes to get there. The first is easy. The vertical component of the velocity is equal to zero at point h. This is the point where the upward motion is stopped and the projectile begins to fall back to Earth.

The initial vertical velocity is

v0y = v0·sin

v0y = 150 m/s · sin(45°)

v0y = 106.1 m/s

Now we know the beginning and final velocity. The next thing we need is the acceleration.

The only force acting on the projectile is the force of gravity. Gravity has a magnitude of g and a direction in the negative y direction.

F = ma = -mg

solve for a

a = -g

Now we have enough information to find the time. We know the initial vertical velocity (V0y) and the final vertical velocity at h (vhy = 0)

vhy – v0y = at

0 – v0y = -9.8 m/s2·t

0 – 106.1 m/s = -9.8 m/s2·t

Solve for t

projectile motion math step 3

t = 10.8 s

Now solve the first equation for h

h = v0yt + ½at2

h = (106.1 m/s)(10.8 s) + ½(-9.8 m/s2)(10.8 s)2

h = 1145.9 m – 571.5 m

h = 574.4 m

The highest height the projectile reaches is 574.4 meters.

Added 14 days ago|2/9/2024 12:48:37 PM

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B.) We’ve already done most of the work to get this part of the question if you stop to think. The projectile’s trip can be broken into two parts: going up and coming down.

ttotal = tup + tdown

The same acceleration force acts on the projectile in both directions. The time down takes the same amount of time it took to go up.

tup = tdown

or

ttotal = 2 tup

we found tup in Part a of the problem: 10.8 seconds

ttotal = 2 (10.8 s)

ttotal = 21.6 s

The total time aloft for the projectile is 21.6 seconds.

ttotal = tup + tdown

The same acceleration force acts on the projectile in both directions. The time down takes the same amount of time it took to go up.

tup = tdown

or

ttotal = 2 tup

we found tup in Part a of the problem: 10.8 seconds

ttotal = 2 (10.8 s)

ttotal = 21.6 s

The total time aloft for the projectile is 21.6 seconds.

Added 14 days ago|2/9/2024 12:49:03 PM

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C.) To find the range, we need to know the initial velocity in the x direction.

v0x = v0cos

v0x = 150 m/s·cos(45)

v0x = 106.1 m/s

To find the range R, use the equation:

R = v0xt + ½at2

There is no force acting along the x-axis. This means the acceleration in the x-direction is zero. The equation of motion is reduced to:

R = v0xt + ½(0)t2

R = v0xt

The range is the point where the projectile strikes the ground which happens at the time we found in Part b of the problem.

R = 106.1 m/s · 21.6s

R = 2291.8 m

v0x = v0cos

v0x = 150 m/s·cos(45)

v0x = 106.1 m/s

To find the range R, use the equation:

R = v0xt + ½at2

There is no force acting along the x-axis. This means the acceleration in the x-direction is zero. The equation of motion is reduced to:

R = v0xt + ½(0)t2

R = v0xt

The range is the point where the projectile strikes the ground which happens at the time we found in Part b of the problem.

R = 106.1 m/s · 21.6s

R = 2291.8 m

Added 14 days ago|2/9/2024 12:49:17 PM

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D.)

The position has two components: horizontal and vertical position. The horizontal position, x, is far downrange the projectile is after firing and the vertical component is the current altitude, y, of the projectile.

To find these positions, we will use the same equation:

d = v0t + ½at2

First, let’s do the horizontal position. There is no acceleration in the horizontal direction so the second half of the equation is zero, just like in Part c.

x = v0xt

We are given t = 10 seconds. V0x was calculated in Part c of the problem.

x = 106.1 m/s · 10 s

x = 1061 m

Now do the same thing for the vertical position.

y = v0yt + ½at2

We saw in Part b that v0y = 109.6 m/s and a = -g = -9.8 m/s2. At t = 10 s:

y = 106.1 m/s · 10 s + ½(-9.8 m/s2)(10 s)2

y = 1061 – 490 m

y = 571 m

At t=10 seconds, the projectile is at (1061 m, 571 m) or 1061 m downrange and at an altitude of 571 meters.

The position has two components: horizontal and vertical position. The horizontal position, x, is far downrange the projectile is after firing and the vertical component is the current altitude, y, of the projectile.

To find these positions, we will use the same equation:

d = v0t + ½at2

First, let’s do the horizontal position. There is no acceleration in the horizontal direction so the second half of the equation is zero, just like in Part c.

x = v0xt

We are given t = 10 seconds. V0x was calculated in Part c of the problem.

x = 106.1 m/s · 10 s

x = 1061 m

Now do the same thing for the vertical position.

y = v0yt + ½at2

We saw in Part b that v0y = 109.6 m/s and a = -g = -9.8 m/s2. At t = 10 s:

y = 106.1 m/s · 10 s + ½(-9.8 m/s2)(10 s)2

y = 1061 – 490 m

y = 571 m

At t=10 seconds, the projectile is at (1061 m, 571 m) or 1061 m downrange and at an altitude of 571 meters.

Added 14 days ago|2/9/2024 12:49:42 PM

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