Show that the path followed by a projected horizontally frop top of the tower with uniform speed is a parabola .​

Explained below.

Explanation:

Let the initial velocity from the top of the tower of height(h) be u'

Now, since it is projected horizontally, the horizontal distance covered will be due to having a uniform horizontal velocity (u) while it will possess a vertical distance in the downward direction due to constant acceleration due to gravity (g).

If the time it takes the ball to reach the ground is "t", then we can say the horizontal distance travelled by the ball is denoted as; x(t) = x, while the vertical distance is denoted by; y(t) = y

Now, since it's a projectile the intial velocity (u) will have vertical and horizontal components which are;

u_y and u_x respectively.

Applying kinematic equations, we have;

x = u_x•t + ½at²

Acceleration is zero in the horizontal x direction. Thus;

x = u_x•t

For the vertical y-direction;

y = u_y•t + ½at²

Here since direction is in that of gravity, then a = g.

Also, since the initial velocity has no downward component, then u_y = 0 m/s

Thus;

y = ½gt²

From x = u_x•t, we have;

t = x/u_x

Thus;

y = ½g(x/u_x)²

y = ½gx²/(u_x)²

Let g/(u_x)² be treated as a constant with the letter k.

Thus;

y = kx²

This is the form of a parabolic equation.

Thus, it has been proved that the projectile follows the path of a parabola.

this is an example of "committed" time.

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