A ball is thrown horizontally from the top of a building 150 meters high. The ball strikes the ground 56 meters horizontally from the point of release. What is the speed of the ball just before it strikes the ground?
To find the speed of the ball just before it strikes the ground, we can use the principle of projectile motion. Since the ball is thrown horizontally, there is no vertical velocity. Therefore, the only force acting on the ball is the force of gravity in the vertical direction.
The time it takes for the ball to hit the ground horizontally can be found using the horizontal distance traveled by the ball. In this case, the horizontal distance is given as 56 meters. We can use the formula:
distance = velocity * time
Since the initial horizontal velocity is constant throughout the motion, we can find the time using the horizontal distance and the velocity of the ball.
Now, let's focus on the vertical motion. The ball is dropped from a height of 150 meters and falls under the influence of gravity. We can use the equation of motion in vertical direction:
distance = initial velocity * time + (1/2) * acceleration * time^2
Here, the initial vertical velocity is zero since the ball was thrown horizontally. The acceleration due to gravity is approximately 9.8 m/s^2.
Since the ball strikes the ground, the distance it traveled vertically is equal to the initial height of the building, which is 150 meters.
Now we can substitute the values into the equation:
150 = 0 * t + (1/2) * 9.8 * t^2
150 = 4.9 * t^2
Divide both sides by 4.9:
t^2 = 30.6122449
Taking the square root of both sides:
t = 5.527 secs (approximately)
Now, let's use the horizontal distance and the time to find the horizontal velocity:
56 = v * 5.527
v ≈ 10.12 m/s
Therefore, the speed of the ball just before it strikes the ground is approximately 10.12 m/s.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*150 = 2,940
V = 54.2 m/s.