Introduction:

The projectile is an object on which the single force acting is gravity. There are a variety of examples of projectiles. The object dropped from rest is the projectile (given that influence of air resistance is minor). The object which is thrown vertically upward is also projectile (given that influence of air resistance is negligible). And the object that is thrown upward at the angle to horizontal is also a projectile (given that influence of air resistance is negligible). The projectile is any object which once projected or dropped goes on in motion by its own inertia and is influenced only by downward force of gravity.

Trajectory:

We can evaluate trajectory of the ball undergoing projectiles motion by plotting height y versus its x-position. We know both x and y as functions of time, and we can eliminate time dependence by using suitable equations of motion. Thus from

x = 0 + (v₀cos Θ₀)t+1/2(0)t²

x = v₀cos Θ₀t

Therefore t = x/( v₀cos Θ₀)

Now using the equation for y

Y = 0 + (v₀sin Θ₀)t - 1/2 gt² = (v₀sin Θ₀)t - 1/2 gt²

Y becomes after substituting for t

Y = 0 + ((v₀sin Θ₀) x/(v₀cos Θ₀) - 1/2g(x/((v₀cosΘ₀)²

i.e.

y = (tan Θ₀)x(-g/(2v₀² cos² Θ₀)x²

We see that coefficient of x and x² are both constants, so this equation have the form:

y = c₁x - c₂ x²

This is a general equation of the parabola. Therefore we conclude that trajectory of all objects moving with the constant acceleration is parabolic. So, plotting different values of x, with the corresponding values of y will trace trajectory of the projectile.

The Flight Time:

At the top of trajectory y-velocity of the projectile will be 0.0 m/s. Object is still moving at this moment, but the velocity is entirely horizontal. At the top, for the time, it has stop moving up and is set to start moving down. At that moment projectile is not moving up nor down, only across.

At the top:

Two dimensional velocity:  Non-zero, aimed to right

X-velocity:  Positive non-zero, equal to two dimensional velocity

Y-velocity:   0.0 m/s

At this point its motion is horizontal; i.e. the vertical component of velocity is zero. This occurs at time t = T/2. Now, we can find T/2 by putting V_y = 0 in the following Eqn.

V_y = v₀sin Θ₀ -gt

becomes,

0 = v₀sin Θ₀ -gt

So far t= T/2

0 = v₀sin Θ₀ -gT/2

Therefore T = (2v₀/g)(sin Θ₀)

Range:

If you shoot the projectile at the angle, when we compute projectile motion, set aside horizontal and vertical components of motion. This is due to force of gravity only applies on projectile in vertical direction, and horizontal component of trajectory's velocity remains uniform.

Imagine that we fire the cannonball at the angle. Given initial speed of cannonball and angle at which it was shot. To determine how far it will travel break the initial velocity into x and y components:

V_xi = V_i cos Θ

 V_yi = V_i sin Θ

The given velocity components are independent and gravity applies only in y direction, which signifies that v_x is constant; only v_y changes with time, utilizing the given equation:

V_y = V_yi + at, or

V_y =  V_i sin Θ - gt

If you want to know the x and y positions of cannonball at any time, you can easily estimate them. You know that x is:

X = V_xt=( V_i cosΘ) t

And as gravity accelerates cannonball vertically, here's what y appears (t² here is what provides cannonball's trajectory in its parabolic shape):

Y = V_{yi t-1/2gt}²

You solve time it takes a cannonball to hit ground when shot straight up (ignoring air resistance) like:

V_f =  V_i + at

0 = V_i - gt    (at maximum height, V_f =0; a = -g = -9.8m/s²)

t_top = V_i/g

t_bottom = 2V_y/g

Knowing time permits you to determine range of cannon in x direction:

S = V_xt_bottom = 2V_xV_y/g = 2V_i²sinΘcosΘ)/g

Maximum Height reached: H = V₀²sin²Θ₀/2g

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