A ball is thrown horizontally from the building roof at a speed of 20 m/s. Takes ball 4 seconds to reach the ground. How tall is building?
What is the ball’s speed and direction as it contacts the ground?
How far from the base of the building does the ball land?
Use conservation of energy to determine the speed of the ball as it contacts the ground.
h = 0.5g*t^2 = = 4.9*4^2 = 78.4 m.
Y = Yo + g*t = 0 + 9.8*4 = 39.2 m/s.
V = Xo + Y = 20 + 39.2i = 44m/s[63o].
d = Xo*t = 20 * 4 = 80 m. from base of bldg.
To find the height of the building, we can use the equation of motion for an object in free fall. The equation is given by:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes for the object to fall.
In this case, since the ball takes 4 seconds to reach the ground, we can plug in the values into the equation:
h = (1/2) * 9.8 * (4)^2
= 78.4 meters
Thus, the height of the building is 78.4 meters.
To determine the ball's speed and direction as it contacts the ground, we can use the fact that the ball is thrown horizontally. Since there is no vertical component to the ball's initial velocity, its horizontal speed remains constant throughout its motion. Therefore, the ball's speed as it contacts the ground is 20 m/s, and its direction is horizontal.
To calculate how far from the base of the building the ball lands, we can use the formula:
d = v * t
where d is the distance, v is the horizontal velocity of the ball, and t is the time of flight.
In this case, the horizontal velocity (v) is equal to the initial horizontal speed of the ball, which is 20 m/s, and the time of flight (t) is 4 seconds. We can substitute these values into the formula:
d = 20 * 4
= 80 meters
Thus, the ball lands 80 meters away from the base of the building.
To determine the speed of the ball as it contacts the ground using conservation of energy, we can consider the initial and final mechanical energy of the ball. The initial mechanical energy is given by the sum of the kinetic energy (0.5 * m * v^2) and the potential energy (m * g * h), where m is the mass of the ball, v is the initial velocity, g is the acceleration due to gravity, and h is the height from which the ball is thrown. Since the ball is thrown horizontally, it does not have any initial vertical velocity and hence no initial potential energy.
Therefore, the initial mechanical energy is only due to the kinetic energy and can be written as:
Initial mechanical energy = 0.5 * m * v^2
At the moment of contact with the ground, all of the initial mechanical energy is converted into kinetic energy, since the potential energy is zero. Therefore, the kinetic energy of the ball as it contacts the ground can be written as:
Kinetic energy = 0.5 * m * (velocity at contact)^2
Since the initial mechanical energy and the kinetic energy at contact are equal, we can equate these two expressions and solve for the velocity at contact:
0.5 * m * v^2 = 0.5 * m * (velocity at contact)^2
Simplifying and solving for the velocity at contact:
v^2 = (velocity at contact)^2
Taking the square root of both sides:
v = velocity at contact
Therefore, the speed of the ball as it contacts the ground using conservation of energy is equal to the initial speed of the ball, which is 20 m/s.