Projectile Motion Simulation-cannonball


Quiz1. The velocity of a typical projectile can be represented by horizontal and vertical components. Assuming negligible air resistance, the horizontal component along the path of the projectile

A. increases.
B. decreases.
C. remains the same.
D. not enough information

Explanation: C
Since there is no force horizontally, no horizontal acceleration occurs.

Quiz2. A ball tossed at an angle of 30 with the horizontal will go as far downrange as one that is tossed at the same speed at an angle of

A. 45.
B. 60.
C. 75.
D. none of the above

Explanation: B
Same initial-speed projectiles have the same range when their launching angles add up to 90. Why this is true involves a bit of trigonometry—which, in the interest of time, we’ll not pursue here.

Projectile Motion Simulation

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Consider the motion of a ball thrown up raised obliquely in the direction of the angle θ with respect to the horizontal direction. In this case, the first velocity v0 is the case that the ingredients in addition to the vertical horizontal component in the horizontal direction differs from the thrown. The movement of such an object, it is convenient to think of it decomposed in the vertical direction and the horizontal direction. An equivalent linear speed movement, such as movement of the ball thrown raised straight up for the vertical direction, because the constant-speed linear motion may be horizontal direction for describing the motion of the moving object by combining the two.
The origin of the coordinate as a starting point, as shown above, and the horizontal x-axis, assuming a vertical upwards in the y-axis gravity acceleration g is the y-axis negative (-) and the direction of the initial velocity v0  is decomposed in the x, y components it can be considered.

Horizontal direction, and acceleration, and the first speed in the vertical direction are represented as follows.

$\ a_x = 0$, $\ a_ y = g$
$\ v_{0x} = v_0 cos\theta$,  $\ v_{0y} = v_0 sin\theta$

The starting time t = 0, an arbitrary time point $\ t$ that the position of the ball in the P(x, y) velocity components at the point P $\ v_x , v_y, a$ displacement component x, y horizontal Since the constant-speed linear motion direction

$\ a_x = 0$
$\ v_x = v_{0x} = v_0 cos\theta$
$\ x = v_{0x}t = (v_0 cos\theta)t$

Because the vertical direction is equivalent to a linear motion speed
$\ a_y = -g$
$\ v_y = v_{0y} - gt = (v_0 sin\theta) - gt$
$\ y = v_{0y}t - \frac{1}{2}at^2= (v_0 sin\theta)t - \frac{1}{2}at^2 $

In addition, the peak $\ v_y = 0$ , so the object is to climb the peak time $\ t_H$ in formula (I. 1-19)
$\ t_H = \frac{v_{0} sin\theta}{g}$

And substituting this in formula (I. 1-20) peak height H is as follows.
$\ H = \frac{v_{0}^2 sin^2\theta}{2g}$

An object is the time to reach the ground $\ t_R$ is $\ y = 0$ Substituting in the expression 0 (I. 1-20)
$ \ t_R = \frac{2v_0 sin\theta}{g}$

And substituting this in formula (I. 1-17) reaches a horizontal distance R is
$\ R=v_0 cos\theta \times \frac {2v_0 sin\theta}{g} =  \frac{v_{0}^2 sin2\theta}{g}$

It becomes. Therefore, when the horizontal reach 45° with respect to one direction of the horizontal plane of the first velocity distance R can be seen that the maximum.

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