A ball is thrown vertically downwards with an initial speed of 4.0 m s-1. The ball hits the groundwith a speed of 16 m s-1. Air resistance is negligible. What is the time of fall and what is thedistance travelled by the ball?
Question
A ball is thrown vertically downwards with an initial speed of 4.0 m s-1. The ball hits the groundwith a speed of 16 m s-1. Air resistance is negligible. What is the time of fall and what is thedistance travelled by the ball?
Solution
To solve this problem, we can use the equations of motion.
- First, let's find the time of fall. We can use the equation v = u + gt, where:
- v is the final velocity (16 m/s)
- u is the initial velocity (4 m/s)
- g is the acceleration due to gravity (approx. 9.8 m/s²)
- t is the time we want to find
Rearranging the equation to solve for t gives us t = (v - u) / g. Substituting the given values gives us t = (16 m/s - 4 m/s) / 9.8 m/s² = 1.22 seconds.
- Next, let's find the distance travelled by the ball. We can use the equation s = ut + 0.5gt², where:
- s is the distance we want to find
- u is the initial velocity (4 m/s)
- t is the time we just found (1.22 seconds)
- g is the acceleration due to gravity (approx. 9.8 m/s²)
Substituting the given values gives us s = 4 m/s * 1.22 s + 0.5 * 9.8 m/s² * (1.22 s)² = 4.88 m + 7.29 m = 12.17 meters.
So, the ball falls for approximately 1.22 seconds and travels approximately 12.17 meters.
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