Part 1:

A ball is thrown horizontally from the top of a building 28.6 m high. The ball strikes the ground at a point 80.3 m from the base of the building.

The acceleration of gravity is 9.8 m/s^2. Find the time the ball is in motion. Answer in units of s.

Part 2:
Find the initial velocity of the ball. Answer in units of m/s.

Part 3:
Find the x componet of its velocity just before it strikes the ground. Answer in units of m/s.

Part 4:
Find the y componet of its velocity just before it strikes the ground. Answer in units of m/s.

Sound with a frequency of 261.6Hz travels through water at a speed of 1435m/s.

Part 1:

To find the time the ball is in motion, we can use the equation of motion for vertical motion:

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

where:
y is the vertical displacement (28.6 m, the height of the building),
u is the initial vertical velocity (which is 0 since the ball is thrown horizontally),
t is the time, and
a is the acceleration due to gravity (-9.8 m/s^2, since it acts downwards).

Plugging in the values, we get:

28.6 = 0*t + (1/2)*(-9.8)*t^2

Rearranging the equation, we have a quadratic equation:

4.9*t^2 = 28.6

Dividing both sides by 4.9:

t^2 = 5.83673469

Taking the square root of both sides:

t ≈ 2.416 s

So, the time the ball is in motion is approximately 2.416 s.

Part 2:
To find the initial velocity of the ball, we can use the horizontal distance traveled (80.3 m) and the time of motion (2.416 s) we just calculated. The equation we can use is:

v = d/t

where:
v is the horizontal velocity (which is also the initial velocity),
d is the horizontal distance, and
t is the time.

Plugging in the values, we get:

v = 80.3/2.416

v ≈ 33.27 m/s

So, the initial velocity of the ball is approximately 33.27 m/s.

Part 3:
To find the x-component of the velocity just before the ball strikes the ground, we can use the equation:

v_x = v_initial

Since the initial horizontal velocity of the ball remains constant throughout the motion, the x-component of its velocity just before it strikes the ground will be the same as the initial horizontal velocity.

Therefore, the x-component of the velocity just before the ball strikes the ground is approximately 33.27 m/s.

Part 4:
To find the y-component of the velocity just before the ball strikes the ground, we can use the equation of motion for vertical motion:

v_y = u + a*t

where:
v_y is the vertical component of the velocity,
u is the initial vertical velocity (which is 0 since the ball is thrown horizontally),
a is the acceleration due to gravity (-9.8 m/s^2), and
t is the time.

Plugging in the values, we get:

v_y = 0 + (-9.8)*(2.416)

v_y ≈ -23.66 m/s

So, the y-component of the velocity just before the ball strikes the ground is approximately -23.66 m/s.