A ball Is thrown horizontally from the top of a 20m high cliff. If the velocity is 8m/s. Calculate;

Question

1. How far from the foot of the cliff did the ball strike the ground

Answer 1

To find out how far from the foot of the cliff did the ball strike the ground, we can use the equation of motion:

s = u * t + (1/2) * a * t^2

where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.

In this problem, the ball is thrown horizontally, so its initial vertical velocity is 0m/s. Therefore, the only vertical force acting on the ball is gravity, causing it to accelerate downwards at a rate of 9.8m/s^2.

Using the equation of motion for the vertical motion:

s = (1/2) * a * t^2

where s is the vertical distance, a is the acceleration (9.8m/s^2 in this case), and t is the time.

Since we know the initial vertical velocity is 0m/s and the vertical distance traveled is 20m, we can use the equation to solve for the time:

20m = (1/2) * 9.8m/s^2 * t^2

Simplifying the equation, we get:

t^2 = 20m * 2 / 9.8m/s^2
t^2 = 4.08s^2

Taking the square root of both sides, we find:

t ≈ 2.02s

Now that we know the time it takes for the ball to hit the ground is approximately 2.02 seconds, we can calculate the horizontal distance using the equation:

s = u * t

s = 8m/s * 2.02s

s ≈ 16.16m

Therefore, the ball struck the ground approximately 16.16 meters from the foot of the cliff.