Projectiles

The problem of the ‘Monkey and the Hunter’ is a famous one in physics. Here’s the picture.

Where should the hunter aim to be sure of hitting the monkey, at him, above him or below him?

When the monkey sees the muzzle flash from the hunter’s gun, he lets go of the branch, but since both he and the bullet are subject to the same gravitational acceleration, he falls the same vertical distance as the bullet in the same time, more or less, so he’ll get shot if the gun is aimed directly at him.

Here’s a little game to play Try to shoot the monkey.

The first successful attempt to describe projectile motion quantitatively followed from Galileo’s insight that the horizontal and vertical motions should be considered separately.

Drop two coins of different mass from the same height. Do they hit the ground together? Why?

Push a coin off the side of the table and drop another from the same height. Do they hit the ground at the same time?

Projectile motion can be described by putting these together.

Galileo argued that, if air resistance could be neglected, the horizontal motion was one at constant velocity, the vertical motion was one of constant acceleration, identical to that of an object falling straight down. Put another way, vertical motion is about the equations of motion, horizontal motion is about speed=distance/time, with a=0. We  illustrate this with an example from history.

The English archers won the Battle of Agincourt in 1415 because the arrows’ release speed hence range of the better-made and longer English bows was greater than the bows used by their French enemies. They could probably count on a release speed of 100 feet per second = 3000 cm/s = 30m/s.

So 30 cos 45 = horizontal speed = 21.2m/s. Using equations of motion with 30sin45 as u, and acceleration  = g, yields t = 2.16s going up, same coming down so total time – 4.3(3)s. So, range =42.4×4.33 = 183m…as long as air resistance is neglected.

Imagine kicking a rugby ball. Suppose the ball leaves the kicker’s boot with a speed of 20ms-1 at 450 to the horizontal. We can use the equations of motion to find out how high it reaches and speed=distance/time to find out how far it travels horizontally, called the Range.

• The initial velocity, u=20sin 450
• The final velocity is zero at the maximum height
• Acceleration  = g
• Horizontal velocity = constant with time at 20cos450

Plotting a graph of vertical displacement against horizontal distance shows us the trajectory of the rugby ball.

You will see this spreadsheet in class. NB: for MAX range, take-off angle = 450.