Learning Goal: To understand the basic concepts of projectile motion. Projectile

Question

Learning Goal:

To understand the basic concepts of projectile motion.

Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really little different from the one-dimensional motions that you may already have studied.

One of the most often used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity v? . In general, this velocity can point in any direction in the xyplane and can have any magnitude. To make a problem more manageable, it is common to break up such a quantity into its x component v? x and itsy component v? y.

*Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed vx must the ball on the left start with so that it hits the ground at the same position as the ball on the right?

Explanation / Answer

Consider a particle with initial velocity that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

What is the x component Vx of V?
What is the y component Vy of V?
How long t(sub g) does it take for the balls to reach the ground? Use 10 m/s^2 for the magnitude of the acceleration due to gravity.

V = 12.0 m/s angle 60 degs above x axis

Vx/V = -cos 60
Vx = -Vcos60 = -0.5*V = -6.0 m/s
I took positive x as positive velocity

Vy/V = sin 60
Vy = Vsin60 = 12.0 * 0.8660 = 10.4 m/s
I took positive y as positive velocity

a = 10 m/s^2

V = u + at
0 = 10.4 - 10t
10t = 10.4
t = 10.4/10 = 1.04 seconds
that means it takes 1.04 seconds to reach the peak

so it would take 2 times that to reach the ground again.
time = 2*1.04 = 2.08 seconds

I hope that helped

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