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?

See previous 5:35 PM post.

To determine the speed of the ball just before it strikes the ground, we can use the equation of motion for an object in free fall:

s = ut + 1/2 * a * t^2

where:
s = distance traveled (in this case, the height of the building)
u = initial velocity (in this case, the horizontal component of the velocity of the ball)
a = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken to reach the ground

Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s. Therefore, the initial velocity of the ball (u) is the same as its horizontal velocity.

First, let's find the time taken for the ball to reach the ground. We'll use the horizontal distance traveled by the ball (56 meters) and the horizontal component of the velocity (u):

s = ut
56 = u * t (equation 1)

Next, let's find the height of the building (s) traveled by the ball:

s = 1/2 * a * t^2
150 = 1/2 * 9.8 * t^2

Now we have two equations with two unknowns (u and t). We can solve these equations simultaneously to find the values of u and t.

From equation 1, rearrange it to solve for u:

u = 56 / t (equation 2)

Substitute the value of u from equation 2 into the height equation:

150 = 1/2 * 9.8 * t^2

Rearrange this equation to solve for t:

4.9 * t^2 = 150
t^2 = 30.6122
t ≈ 5.52 seconds

Now substitute this value of t into equation 2 to solve for u:

u = 56 / t
u = 56 / 5.52
u ≈ 10.14 m/s

Therefore, the speed of the ball just before it strikes the ground is approximately 10.14 m/s.