Two baseballs are thrown off the top of a building that is 7.24 m high. Both are thrown with initial speed of 63.3 mph. Ball 1 is thrown horizontally, and ball 2 is thrown straight down. What is the difference in the speeds of the two balls when they touch the ground? (Neglect air resistance.)
vball 1 - vball 2 = 1
A ball is thrown upward from the ground with an initial speed of 26.8 m/s; at the same instant, a ball is dropped from a building 15.9 m high. After how long will the balls be at the same height?
To find the difference in speeds between ball 1 and ball 2 when they touch the ground, we need to calculate their final velocities.
First, let's find the final velocity of ball 1 when it touches the ground. Since ball 1 is thrown horizontally, it only has an initial horizontal velocity and no initial vertical velocity. We can use the following kinematic equation to find the time it takes for the ball to reach the ground:
h = (1/2) * g * t^2
Where:
h is the initial height of the building (7.24 m)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken to reach the ground
Rearranging the equation to solve for t:
t = sqrt((2h) / g)
Plugging in the values, we get:
t = sqrt((2 * 7.24) / 9.8) ≈ 1.07 seconds
Since ball 1 has no vertical velocity, its final vertical velocity will also be zero when it touches the ground. Therefore, its final velocity will only have a horizontal component, which is equal to its initial horizontal velocity.
Now, let's find the final velocity of ball 2 when it touches the ground. Since ball 2 is thrown straight down, we can use the following kinematic equation to find the time it takes for the ball to reach the ground:
h = (1/2) * g * t^2
Rearranging the equation to solve for t:
t = sqrt((2h) / g)
Plugging in the values, we get:
t = sqrt((2 * 7.24) / 9.8) ≈ 1.07 seconds
Just like ball 1, ball 2 will have a final vertical velocity of zero when it touches the ground, but its final velocity will have a downward vertical component.
Since both balls take the same amount of time to reach the ground, their time of flight is the same. Therefore, the difference in their speeds will only be determined by their vertical components of velocity.
To find the difference in their final velocities, we need to calculate the magnitude of the vertical velocities of ball 1 and ball 2.
For ball 1, we know that the initial speed is 63.3 mph (miles per hour). We need to convert this to m/s:
63.3 mph = 63.3 * 0.447 m/s ≈ 28.28 m/s
Since ball 1 has no vertical component of velocity, its vertical velocity will be 0 m/s.
For ball 2, since it is thrown straight down, its initial vertical velocity is equal to its initial speed:
v2 = 63.3 * 0.447 m/s ≈ 28.28 m/s
Therefore, the difference in speeds between ball 1 and ball 2 when they touch the ground is:
vball 1 - vball 2 = 0 m/s - 28.28 m/s = -28.28 m/s
The negative sign indicates that ball 2 has a higher speed than ball 1 when they touch the ground, due to the downward vertical component of velocity for ball 2.