Four balls are taken to the top of a building. Ball 1 is dropped (zero initial velocity). Ball 2 is thrown straight up with a velocity of 3 m/s. Ball 3 is thrown straight out with a horizontal velocity of 3 m/s. Ball 4 is thrown straight down with a velocity of 3 m/s. Which ball hits the ground first?
Match the projectile paths with the described motions.
A. A tennis ball is thrown straight up.
B. A golf ball is hit off of the tee.
C. A rock is thrown horizontally off a cliff.
D. An apple falls from a tree.
To determine which ball hits the ground first, we need to analyze the motion of each ball separately. We'll consider the effects of gravity on the vertical motion of the balls.
Ball 1 is dropped from the top of the building, so it starts with an initial velocity of 0 m/s. Since only gravity affects its motion, it will accelerate downwards at a rate of 9.8 m/s².
Ball 2 is thrown straight up with an initial velocity of 3 m/s. Due to gravity, it will experience negative acceleration and gradually lose its upward velocity. Eventually, it will reach its highest point and start coming back down. The time it takes for Ball 2 to reach its peak and come back down can be determined by using the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the final velocity is 0 m/s (at the highest point), the initial velocity is 3 m/s, and the acceleration is -9.8 m/s² (due to gravity).
0 = 3 + (-9.8)t,
-3 = -9.8t,
t ≈ 0.31 seconds.
Therefore, Ball 2 will take approximately 0.31 seconds to reach the highest point. Since the vertical motion is symmetric, it will also take 0.31 seconds to come back down to the ground.
Ball 3 is thrown straight out with a horizontal velocity of 3 m/s. Since there is no vertical component to its initial velocity, it will not be affected by gravity in the vertical direction. It will continue to move horizontally with constant velocity and never hit the ground.
Ball 4 is thrown straight down with an initial velocity of 3 m/s. Similar to Ball 1, it will be affected by gravity and accelerate downwards at 9.8 m/s². However, it starts with an initial velocity, so it will initially take longer to reach the ground compared to Ball 1.
To find the time it takes for Ball 4 to hit the ground, we can use the equation:
s = ut + (1/2)at²,
where s is the vertical distance covered, u is the initial velocity, a is the acceleration, and t is the time. In this case, the vertical distance covered (s) is not given, but we can assume it to be the same height as the building (we'll use h for the height).
h = 3t + (1/2)(-9.8)t²,
0 = 3t - (4.9)t²,
4.9t² - 3t = 0,
t(4.9t - 3) = 0.
This equation has two solutions: t = 0, and t ≈ 0.61 seconds. The solution t = 0 corresponds to the initial time when the ball is thrown. Since we are interested in the time it takes for the ball to hit the ground, the relevant solution is t ≈ 0.61 seconds.
Therefore, Ball 4 will take approximately 0.61 seconds to hit the ground.
In summary, Ball 1 (dropped ball) will hit the ground first, followed by Ball 4 (thrown straight down), and Ball 2 (thrown straight up) will take the longest time to come back down. Ball 3 (thrown straight out) will never hit the ground if there is no vertical component to its velocity.