A ball is thrown horizontally from the top of a cliff 20m high. If the initial horizontal velocity is 8.0m/s,find how long it takes to reach the horizontal plane at the foot of the cliff.

Let u be the horizontal speed and v the vertical.

u(0)=8.0m/s, v(0)=0m/s

Let x be the horizontal height and y the vertical.
x(0)=0m, y(0)=20m

y(t) = y(0) + v(0) t -(9.81/2) t^2

Find t when y(t) = 0m
So: 0m = 20m - 4.905(m/s^2) t^2

From equation of motion. V^2=U^2+2gh find v. Applying the parameter given v=21m/s. From a=v-u/t then substitute for v,u and g given to find t. T=1.36sec

Formula=V=U +or- gt

You remove( U+or_) because you are given initial horizontal velocity not initial velocity.
So you have (V=gt)
Because you are looking for (t)
You make it the subject of formula,you now have T=V÷g
Then you apply the numbers
T=8÷10
=O.8s(seconds)

To find the time it takes for the ball to reach the horizontal plane at the foot of the cliff, we can use the equations of motion.

Let's assume that the initial horizontal velocity of the ball is denoted as vₓ and the time it takes to reach the horizontal plane is t.

Since the ball is thrown horizontally, there is no initial vertical velocity (vᵧ = 0). The only force acting on the ball in the vertical direction is gravity. Therefore, we can use the kinematic equation for vertical motion:

h = vᵧt + (1/2)gt²

Where:
h = vertical displacement (20m, since the ball is thrown from the top of a cliff)
vᵧ = initial vertical velocity (0)
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time taken to reach the horizontal plane

Substituting the values we know into the equation, we can solve for t:

20m = 0 + (1/2)(9.8m/s²)t²

Simplifying the equation:

20m = (4.9m/s²)t²

Dividing both sides of the equation by 4.9m/s²:

t² = 20m / 4.9m/s²

t² = 4.08s²

Taking the square root of both sides:

t ≈ √4.08s²

t ≈ 2.02s

Therefore, it will take approximately 2.02 seconds for the ball to reach the horizontal plane at the foot of the cliff.