A ball Is thrown horizontally from the top of a 20m high cliff. If the velocity is 8m/s. Calculate;
1. How far from the foot of the cliff did the ball strike the ground
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.