Question

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground? Neglect air resistance

Answer 1

Answer:

The horizontal velocity of the ball just before it touches the ground is 10.32 m/s

Explanation:

Given;

Horizontal distance of the ball, x = 20m

Vertical displacement of the ball, y = 1.6m

let the horizontal velocity of the ball = vx

The horizontal distance of the ball is given as;

$X = V_x*t\\\\20 = V_x*t-----------(i)$

The vertical displacement of the ball is given as;

$1.6 = V_x*t - \frac{1}{2}gt^2 -----------(ii)$

Substitute in equation (i) into equation (ii)

$1.6 = 20 -\frac{1}{2}gt^2\\\\1.6 = 20 -\frac{1}{2}(9.8)t^2\\\\1.6 -20=-4.9t^2\\\\-18.4 = -4.9t^2\\\\t^2 = \frac{18.4}{4.9} =3.7551\\\\t = \sqrt{3.7551} \\\\t =1.938 sec$

From equation (i), x = Vx*t

Vx = x/t

Vx = 20/1.938

Vx = 10.32 m/s

Therefore, the horizontal velocity of the ball just before it touches the ground is 10.32 m/s

Answer 2

Answer:

10m/s

Explanation:

The motion has both horizontal and vertical components. Using one of the equations of motion as follows;

s = ut + $\frac{1}{2}$(a x t²)          ------------------(i)

Where;

s = vertical / horizontal displacement

u = initial velocity of the body in the vertical or horizontal motion

t = time taken for the displacement

a = acceleration of the body.

Let's consider the horizontal motion first;

Using the same equation (i) as follows;

s = ut + $\frac{1}{2}$(a x t²)            -----------------(ii)

Where;

s = horizontal displacement = 20m

a = acceleration = 0 [acceleration in the horizontal is zero since velocity is constant]

Substitute these values into equation (ii) as follows;

20 = ut + $\frac{1}{2}$(0 x t²)

20 = ut + 0

20 = ut  --------------------------------(iii)

Now let's consider the vertical motion;

Using the same equation (i) as follows;

s = ut + $\frac{1}{2}$(a x t²)            -----------------(iv)

Where;

s = vertical displacement = 1.6m

a = acceleration due to gravity [ since the ball eventually moves downwards in the direction of gravity, the value will be +10m/s²]

u = 0         [since the ball rolls horizontally]

Substitute these values into equation (iv) as follows to get the time taken to touch the ground;

1.6 =  $\frac{1}{2}$(10 x t²)

20 = 5t²

t² = 20 / 5

t² = 4

t = √4

t = 2 seconds.

Now substitute the value of t = 2seconds into equation (iii) above as follows to calculate the initial velocity of the ball;

20 = u (2)

2u = 20

u = 10m/s

Since the initial horizontal velocity (u) is 10m/s, and velocity is constant on the horizontal motion, we can conclude that the horizontal velocity (v) of the ball just before it touches the ground is the same as the initial velocity which is 10m/s