Please can anyone help me with my Projectiles work? I would be grateful if any workings out could be shown.
Cannonballs fired from a sail ship travel a horizontal distance of 192m in 1.6s before striking the sea. Ignore air resistance.
1) Calculate the horizontal velocity of the cannonballs.
2) Determine the height of the cannon above the sea.
3) Calculate the resultant velocity of the cannonballs as they strike the sea.
4) The horizontal velocity is now increased. Explain the effect, if any, that this would have on the time of the flight, i.e. the time taken for the cannonballs to strike the sea.
5) Discuss the effect that air resistance would have on the cannonball's motion.
Any help would be appreciated.
Cannonballs fired from a sail ship travel a horizontal distance of 192m in 1.6s before striking the sea. Ignore air resistance.
1) Calculate the horizontal velocity of the cannonballs.
2) Determine the height of the cannon above the sea.
3) Calculate the resultant velocity of the cannonballs as they strike the sea.
4) The horizontal velocity is now increased. Explain the effect, if any, that this would have on the time of the flight, i.e. the time taken for the cannonballs to strike the sea.
5) Discuss the effect that air resistance would have on the cannonball's motion.
Any help would be appreciated.
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EDIT: sorry... made a few mistakes here...
give me some time to correct it ;-)
1) The horizontal velocity is the horizontal distance, divided by the time it took to travel that distance, so
Vh = 192 / 1.6 = 120 m/s
2) During that 1.6 seconds, the cannonball will also travel a have a vertical displacement. To calculate it, you must first know the vertical component of the cannonballs velocity. You can calculate it using this formula:
Vv = Vi + a * t
where
Vv = final vertical velocity = ?
Vi = initial vertical velocity = 0 m/s
a = acceleration = -9.8 m/s²
t = time = 1.6 s
so
Vv = 0 + (-9.8) * 1.6 = -15.68 m/s
It is negative because it is going downward.
Now you can use this formula to find the vertical distance the cannonball was displaced:
d = ½ * (Vi + Vv) * t
where
d = distance = ?
Vi = initial vertical velocity = 0 m/s
Vv = final vertical velocity = 15.68 m/s
t = time = 1.6 s
so
d = ½ * (0 + 15.68) * 1.6 = 12.54 m
3) To calculate the resultant velocity of the cannonball, you use this formula:
Vr = √((Vh)² + (Vv)²)
where
Vh = horizontal velocity = 120 m/s
Vv = vertical velocity = 15.68 m/s
so
Vr = √((12)² + (15.68)²) = 19.
To calculate the angle at which it hits the water, you use this formula
θ = arctan(dy/dx)
where
dy = vertical distance = 12.54 m
dx = horizontal distance = 120 m
so
θ = arctan(12.54/120) = 5.9°
4) The time the cannonball remains in the air, is only affected by the vertical velocity. Changing the horizontal velocity will only affect the horizontal distance, but the cannonball will hit the water at the exact same time as before.
5) If any air resistance, it will affect both the horizontal and vertical component by slowing it down. The cannonball will hit the water after a shorter time, and it will travel less horizontal distance. If the initial velocity also had an upward going component, then it would go less high.
give me some time to correct it ;-)
1) The horizontal velocity is the horizontal distance, divided by the time it took to travel that distance, so
Vh = 192 / 1.6 = 120 m/s
2) During that 1.6 seconds, the cannonball will also travel a have a vertical displacement. To calculate it, you must first know the vertical component of the cannonballs velocity. You can calculate it using this formula:
Vv = Vi + a * t
where
Vv = final vertical velocity = ?
Vi = initial vertical velocity = 0 m/s
a = acceleration = -9.8 m/s²
t = time = 1.6 s
so
Vv = 0 + (-9.8) * 1.6 = -15.68 m/s
It is negative because it is going downward.
Now you can use this formula to find the vertical distance the cannonball was displaced:
d = ½ * (Vi + Vv) * t
where
d = distance = ?
Vi = initial vertical velocity = 0 m/s
Vv = final vertical velocity = 15.68 m/s
t = time = 1.6 s
so
d = ½ * (0 + 15.68) * 1.6 = 12.54 m
3) To calculate the resultant velocity of the cannonball, you use this formula:
Vr = √((Vh)² + (Vv)²)
where
Vh = horizontal velocity = 120 m/s
Vv = vertical velocity = 15.68 m/s
so
Vr = √((12)² + (15.68)²) = 19.
To calculate the angle at which it hits the water, you use this formula
θ = arctan(dy/dx)
where
dy = vertical distance = 12.54 m
dx = horizontal distance = 120 m
so
θ = arctan(12.54/120) = 5.9°
4) The time the cannonball remains in the air, is only affected by the vertical velocity. Changing the horizontal velocity will only affect the horizontal distance, but the cannonball will hit the water at the exact same time as before.
5) If any air resistance, it will affect both the horizontal and vertical component by slowing it down. The cannonball will hit the water after a shorter time, and it will travel less horizontal distance. If the initial velocity also had an upward going component, then it would go less high.