A ball is projected horizontally from the top of a 92.0-meter high cliff with an initial speed of 19.8m/s. Determine the horizontal displacement.

85.14 m

(y) denotes "in the vertical direction"

(x) denotes "in the horizontal direction"

We are given:

Initial Horizontal velocity of the Ball (u(x)) = 19.8 m/s

Initial height of the ball (s(y)) = 92 m

Initial Vertical velocity of the Ball (u(y)) =  0 m/s

Time taken to reach the ground:

taking downwards direction as positive

Since the horizontal velocity is not opposed by any force, it will be the same until the ball reaches the ground

The vertical velocity will be increasing at a rate of (10 m/s)/s until the ball hits the ground

ay = 10 m/s²

So, while calculating the time. we can just ignore the horizontal velocity

Solving for the time taken:

s(y) = u(y)t + 1/2a(y)t²                                   [second equation of motion]

92 = (0)(t) + 1/2(10)(t)²                               [replacing the variables]

92 = 5t²

t² = 92/5                                            [dividing both sides by 5]

t = √18.4                                            [taking the square root of both sides]

t = 4.3 seconds

So, it took the ball 4.3 seconds to reach the ground

Horizontal Distance travelled by the ball:

We know that the ball will reach the ground in 4.3 seconds

Since the horizontal velocity will not change, the ball will move with a constant velocity of 19.8 m/s in the horizontal direction

Horizontal distance travelled:

s(x) = u(x)t + 1/2a(x)t²                    [second equation of motion]

s(x) = (19.8)(4.3) + 1/2(0)(t)²          [replacing the variables]

s(x) = 85.14 m

Hence, the ball travels 85.14 m horizontally

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.