A ball is thrown horizontally from the top of a 65 m building and lands 105 m from the base of the building. Ignore air resistance, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive x in the direction of the throw.
1. How long is the ball in the air in seconds?
2. What must have been the initial horizontal component of the velocity in m/s?
3.What is the vertical component of the velocity just before the ball hits the ground in m/s?
To find the answers to the questions, we can use the equations of motion for projectiles.
1. How long is the ball in the air in seconds?
To find the time the ball spends in the air, we need to calculate the time it takes for the ball to fall from the top of the building to the ground. We can use the equation:
y = ut + (1/2)gt^2
where:
- y is the vertical displacement (65 m, the height of the building)
- u is the initial vertical velocity (0 m/s as the ball is thrown horizontally)
- t is the time in seconds
- g is the acceleration due to gravity (9.8 m/s^2)
Since the initial vertical velocity is 0, the equation simplifies to:
y = (1/2)gt^2
Rearranging the equation to solve for t:
t^2 = 2y / g
t = sqrt(2y / g)
Plugging in the values:
t = sqrt(2 * 65 / 9.8) = 3.23 seconds
Therefore, the ball is in the air for approximately 3.23 seconds.
2. What must have been the initial horizontal component of the velocity in m/s?
The horizontal component of the velocity remains constant throughout the motion because there is no horizontal force acting on the ball.
We can use the equation:
x = ut
where:
- x is the horizontal displacement (105 m)
- u is the initial horizontal velocity
- t is the time (3.23 seconds, as calculated in the previous question)
Rearranging the equation to solve for u:
u = x / t
Plugging in the values:
u = 105 / 3.23 ≈ 32.5 m/s
Therefore, the initial horizontal component of the velocity was approximately 32.5 m/s.
3. What is the vertical component of the velocity just before the ball hits the ground in m/s?
The vertical component of the velocity can be found using the equation:
vf = vi + gt
where:
- vf is the final vertical velocity just before the ball hits the ground
- vi is the initial vertical velocity (0 m/s)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time (3.23 seconds)
Plugging in the values and calculating:
vf = 0 + (9.8 * 3.23)
≈ 31.7 m/s
Therefore, the vertical component of the velocity just before the ball hits the ground is approximately 31.7 m/s.