Anonymous

# physics problem - projectile motion! thank you?

2. A large cannon is fired from ground level over level ground at an angle of 79o above the horizontal. This initial speed is 993 m/s. Neglecting air resistance, how far will the projectile travel in the horizontal direction before striking the ground?

Relevance

..............._----_ ←at this point Vy = 0

.........._-‾..........‾-_..………..Vx = still constant

...... _-‾................‾-…………Y or H is max.

Vy.../......................\..t = ½ T (half-flight)

'.↑../.........................\

..|./.79º )...........'........\

..|/---→Vx___________\_____ ground level Y = 0, time = T(full flight)

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓…..V = -V(initial)

Most of the above answerers got it right. This solution is also using the classical equations of Linear Motion. I suggest to the asker that he should be familiar with those and any Free-fall and Projectile Motion could be dealt with sufficiently.

1. First we need to find how long the cannonball is in the air, its "hang time" if you will. This span of time is the time it takes for the cannonball to go up, and fall back down. This is completely dependent on the y-component of the velocity, so we have to use some trigonometry to find this. If you drew a right triangle, you'll find that the y-comp onent:

Velocity at y axis = (Initial Velocity)sin(79) = 993sin79 = 974.756 m/s.

2. Now that you have the velocity at the y axis, We can now find the time it takes for the cannon to land. We can use the equation

Final Velocity = Initial Velocity + (Acceleration)(Time)

or

Vf = Vi + at

note that your aceeleration = -g, since it is pulling it downward so:

Vf = Vi - gt.

Isolating t:

Vf + gt = Vi

gt = Vi - Vf

t = (Vi - Vf)/g

We know the Initial speed in the y axis, which is 974.756 m/s. But what about the final velocity? Here's something interesting: When you throw a ball up at say a certain speed 2m/s, when it falls back to your hand (assuming you kept your hand at the same place) it will return to your hand at a speed of -2m/s, the opposite of the initial velocity. So in the case of the cannonball, I can say that the final velocity is -974.756 m/s. I know my Vi, Vf, and g, so I can now solve for time.

t = (974.756 - (-974.756))/9.8 = (974.756 + 974.756)/9.8 = 1949.512/9.8 = 198.9298 seconds.

3. Why do I need to know the hang time? This is because the ball is traveling at the x direction for this much time, and after that time, it stops moving (unless it keeps on rolling, but that's another problem! Let's assume it stays put where it lands.)

Anyway, I can use the equation

speed = distance/time

or

distance = (speed)(time) = vt

I have my time, but I need my velocity. Remember we're looking for the distance at the x-direction, not the y, so we for the speed here, we need to get the x-component of the projectile's velocity. Using trigonometry again:

Velocity at x axis = (Initial Velocity)cos(79) = 993cos79 = 993(0.19) = 189.473 m/s.

So, using this on the previous equation, and the air-time:

distance traveled along the horizontal direction = vt = (189.473)(198.9298) = 37,691.826 meters.

First you have to calculate the vertical and horizontal components of the projectile velocity:

Velocity is 993m/s at angle 79° above horizontal.

Vel vert = 993sin 79°

Vel vert = 974m/s

Vel hor = 993cos 79

Vel hor = 189.47m/s

Time in the air:

The vertical displacement of the projectile is zero. It goes up and then comes down to the same level

s = ut + 0.5*a*t²

Take up as positive:

0 = 974t + 0.5*(-9.8)t²

0 = 974t - 4.9t²

0 = t(974 - 4.9t)

t = 0 or t = 198.77s

The projectile is in the air for 198.77 seconds. How far does it travel horizontally in this time?

distance = hor velocity * time

distance = 189.47*198.77

distance = 37,662m

theta = 79 deg

vo = 993 m/s

Range R = vo^2 sin(2 theta)/g

= 993^2 * sin(2 * 79deg)/9.8

= 993^2 * sin(158deg)/9.8

= 37692 m

Ans: 37692 m

______________________.

• Anonymous