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On his quest to find a shrubbery for the "Knights Who Say Ni!", King Arthur rides his horse off a 15 m cliff and lands 4.3 m away.

What was his original horizontal velocity?

Answer :

Answer:

v = 2.45 m/s

Explanation:

first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:

h = Vi t + (1/2)gt²

where,

h = height of cliff = 15 m

Vi = Initial Vertical Velocity = 0 m/s

t = time taken = ?

g = 9.8 m/s²

Therefore,

15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²

t² = (15 m)/(4.9 m/s²)

t = √3.06 s²

t = 1.75 s

Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,

s = vt

where,

s = horizontal distance covered = 4.3 m

v = original horizontal velocity = ?

Therefore,

4.3 m = v(1.75 s)

v = 4.3 m/1.75 s

v = 2.45 m/s

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Rewritten by : Barada

Final answer:

The original horizontal velocity of King Arthur is 2.39 m/s.

Explanation:

The horizontal velocity of King Arthur can be calculated using the formula:

vx = d / t

Where vx is the horizontal velocity, d is the horizontal distance traveled (4.3 m), and t is the time of flight. Since the vertical motion and horizontal motion are independent of each other, we can ignore the height of the cliff and focus only on the horizontal distance traveled. Therefore, the horizontal velocity of King Arthur is:

vx = 4.3 m / 1.80 s = 2.39 m/s