A tennis ball is struck and departs from the racket horizontally with a speed of 27.3 m/s. The ball hits the court at a horizontal distance of 18.7 from the racket. How far above the court is the tennis ball when it leaves the racket?
Well, let's break this down. We know that the speed of the ball when it leaves the racket horizontally is 27.3 m/s. Now, in order to find out how far above the court the ball is when it leaves the racket, we'll need to use a little bit of physics and some good old-fashioned humor.
You see, when the ball is struck, it experiences a force from the racket that sends it flying. This force causes the ball to travel in a parabolic trajectory, kind of like a really fancy rollercoaster. And just like a rollercoaster, we can use certain equations to figure out how far up the ball goes.
Now, in order to solve this problem, we need to use the kinematic equation for vertical motion, which is given by the equation:
h = v^2 / (2g)
Where h is the height above the court, v is the initial vertical velocity, and g is the acceleration due to gravity. But since the ball is struck horizontally, we can assume that the initial vertical velocity is zero. So, the equation simplifies to:
h = 0^2 / (2g)
Now, let's talk about gravity. You know, that thing that keeps us all grounded and stops us from floating away into space? Yeah, that one. The acceleration due to gravity on Earth is approximately 9.8 m/s^2.
So, now we can plug in the values into our equation:
h = 0 / (2 * 9.8)
And after simplifying, we get:
h = 0 m
So, according to my calculations, when the tennis ball leaves the racket, it is actually at the same height as the court. It's not above or below, just right at court level. I guess you could say the ball is serving some serious truth with its shot placement!
I hope that answers your question, and remember, never underestimate the power of physics and a little bit of humor!
To determine how far above the court the tennis ball is when it leaves the racket, we can use the principles of projectile motion.
The horizontal distance the ball travels (18.7 m) is only affected by the initial horizontal velocity of the ball. Since the ball is struck horizontally, the only horizontal velocity component is the initial speed of 27.3 m/s.
The vertical distance the ball travels depends on the time it takes to reach the ground and the effect of gravity. Since the ball is not initially given an upward or downward velocity, we can assume that the initial vertical velocity is 0 m/s.
Let's use the equation for vertical distance in projectile motion:
d = (1/2) * g * t^2
Where:
d is the vertical distance traveled
g is the acceleration due to gravity
t is the time of flight
Since the vertical distance traveled when the ball hits the ground is the same as the vertical distance traveled when the ball leaves the racket, we can equate the two equations:
(1/2) * g * t^2 = h
Where:
g is the acceleration due to gravity (approximately -9.8 m/s^2)
t is the time of flight
h is the height above the court
Now, we need to find the time of flight (t). Since the horizontal distance traveled is 18.7 m and the horizontal velocity is 27.3 m/s, we can use the formula:
t = d / v
Where:
t is the time of flight
d is the horizontal distance traveled (18.7 m)
v is the horizontal velocity (27.3 m/s)
Substituting the values, we have:
t = 18.7 m / 27.3 m/s
t ≈ 0.685 seconds (rounded to 3 decimal places)
Now, we can substitute this value of t back into the equation to find the height (h):
(1/2) * g * (0.685 s)^2 = h
Simplifying this equation, we get:
h ≈ (1/2) * (-9.8 m/s^2) * (0.468 s^2)
h ≈ -2.667 m^2/s^2 (rounded to 3 decimal places)
Since the height is a vertical measurement, we ignore the negative sign. Therefore, the tennis ball is approximately 2.667 meters above the court when it leaves the racket.