Demarcus929
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A ball is projected horizontally from the top of a 92.0-meter high cliff with an initial speed of 19.8 m/s. Determine: (a) the horizontal displacement and (b) the final speed the instant prior to hitting the ground.

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olemakpadu
The ball has two independent motions; the vertical motion and the horizontal motion.
That is the ball is falling vertically downwards from the cliff and the same time the ball is moving horizontally forward from the cliff.
Note: Initial Vertical Speed is = 0, Initial Horizontal Speed = 19.8 m/s
For the vertical motion: the initial velocity is zero. That is [tex] u_{y} [/tex] = 0.
S = ([tex] u_{y} [/tex])t + (1/2)gt^2.                            where g = acceleration due to gravity = 10 m/s^2
S = Vertical height =  92m
Substituting [tex] u_{y} [/tex] = 0, and  S = 92m, g = 10m/s^2
92 = 0*t  + (1/2)*10*t^2
92 = (1/2)*10*t^2
92 = 5t^2
5t^2 = 92
t^2 = 92/5
t^2 = 18.4    Take square root of both sides
t = [tex] \sqrt{18.4} [/tex]
t = 4.29 seconds, that is the time to reach the ground. Time is same both for horizontal and vertical motion

a) Horizontal Displacement, which is Range R = Horizontal Speed * time
R = [tex] u_{x} [/tex] * time
R = 19.8 * 4.29 = 84.942 m.

b) Final Speed before hitting the ground.
Remember we realized that speed was in two parts: Vertical Speed and Horizontal Speed.
Just before hitting the ground Horizontal speed is same = 19.8 m/s, this is the same as the initial horizontal projected speed. (No acceleration in Horizontal direction so this remains the same)
Just before hitting the ground, [tex] v_{y} = u_{y} + gt[/tex]
At projection Vertical speed is initially zero  [tex] u_{y} = 0 [/tex]
[tex] v_{y} = 0 + 10*4.29 [/tex]  
Time is same both for horizontal and vertical motion
[tex] v_{y} [/tex] = 42.9  m/s
The velocity just before hitting the ground is 19.8 m/s in the horizontal direction and 42.9 m/s in the vertical direction.
These two are at right angles to each other. Therefore the resultant is:
= [tex] \sqrt{19.8^{2}+ 42.9^{2} } [/tex]
= 47.25 m/s.
This is the resultant velocity just before it hits the ground.

Answer:

Part a)

[tex]d = 85.8 m[/tex]

Part b)

[tex]v = 46.8 m/s[/tex]

Explanation:

Part a)

As we know that ball is projected horizontally

so here the vertical speed of the ball is zero

now we have

[tex]y = \frac{1}{2}gt^2[/tex]

now we have

[tex]92 = \frac{1}{2}(9.8)t^2[/tex]

[tex]t = 4.33 s[/tex]

now in the same time the horizontal distance covered by the ball

[tex]d = v_x t[/tex]

[tex]d = 19.8 \times 4.33[/tex]

[tex]d = 85.8 m[/tex]

Part b)

now the horizontal speed of the ball always remains constant

so we have

[tex]v_x = 19.8 m/s[/tex]

now in vertical direction we have

[tex]v_y = v_0 + at[/tex]

[tex]v_y = 0 + 9.8(4.33)[/tex]

[tex]v_y = 42.4 m/s[/tex]

so the final speed of the ball is given as

[tex]v = \sqrt{v_y^2 + v_x^2}[/tex]

[tex]v = \sqrt{42.4^2 + 19.8^2}[/tex]

[tex]v = 46.8 m/s[/tex]

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