A golf ball is dropped from rest from a height of 8.20 m. It hits the pavement, then bounces back up, rising just 5.40 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.
1.5
Elena is incorrect, in that you must take the square root of each thing.
To solve this problem, we will use the kinematic equations of motion. Let's break down the different stages of the ball's motion and calculate the time for each stage.
1. The ball is dropped from rest from a height of 8.20 m. The equation we will use for this stage is:
h = ut + (1/2)gt^2
Where:
- h is the height (8.20 m),
- u is the initial velocity (0 m/s since it is dropped from rest),
- g is the acceleration due to gravity (-9.8 m/s²),
- t is the time.
Rearranging the equation, we get:
(1/2)gt^2 = h
Simplifying further:
t = √(2h/g)
Plugging in the values, we get:
t₁ = √(2 * 8.20 m / 9.8 m/s²)
2. The ball hits the pavement and bounces back up, rising to a height of 5.40 m. During this stage, we need to consider the time it takes for the ball to reach its maximum height. The equation we will use is:
v = u + gt
Where:
- v is the final velocity at the maximum height (0 m/s),
- u is the initial velocity (which we need to find),
- g is the acceleration due to gravity (-9.8 m/s²),
- t is the time.
Rearranging the equation, we get:
u = -gt
Plugging in the values, we get:
u = -9.8 m/s² * t₁
Now, we can calculate the time it takes for the ball to reach its maximum height using the equation:
v = u + gt
Where:
- v is the final velocity at the maximum height (0 m/s),
- u is the initial velocity (-9.8 m/s² * t₁),
- g is the acceleration due to gravity (-9.8 m/s²),
- t is the time.
Rearranging the equation, we get:
t₂ = v / (-g)
Plugging in the values, we get:
t₂ = 0 m/s / (-9.8 m/s²)
3. After reaching its maximum height, the ball falls back down to the pavement and is caught by the boy. We need to calculate the time it takes for the ball to fall from the maximum height to the catch position at 1.40 m above the pavement.
Using the equation of motion:
h = ut + (1/2)gt^2
Where:
- h is the height (5.40 m),
- u is the initial velocity (0 m/s at the maximum height),
- g is the acceleration due to gravity (-9.8 m/s²),
- t is the time.
Rearranging the equation, we get:
(1/2)gt² - 5.40 m = 0
Simplifying further:
t³ - (10.8 m / g) = 0
Solving this cubic equation will give us the time it takes for the ball to reach the catch position.
Finally, we can calculate the total time the ball is in the air by summing up the times obtained from each stage:
Total time = t₁ + t₂ + t₃
h1=g•t1²/2=>
t1=2•h1/g= 2•8.2/9.8=1.8 s.
Since the time of ascening = time of descending,
t2=2•h2/g=1.1 s.
h3=5.4-1.4 =4 m
t3=2•h3/g=2•4/9.8=0.82 s.
t=1.8+1.1+0.82=3.72 s.