During Service, a tennis ball is struck at height, h = 1.95 m, above a tennis court and moves with velocity Vo= Voxi. Assume the ball encounters no air resistance, and use a Cartesian coordinate system with the origin located at the balls initial position. Vox = 24.5 m/s
a) Calculate the tennis balls horizontal velocity, Vfx in m/s, when it strikes the court.
b) How long, tf, in seconds does the tennis ball move before striking the court?
c) How far horizontally from the service position, xf, in meters does the tennis ball move before striking the court?
Please show me what formulas you will use to solve this because I don't even know how to start calculating it.
Thanks!!
There are no horizontal forces.
Therefore there is NO change in horizontal velocity !!!
Now the vertical problem is another matter entirely.
The ball has no initial speed up or down
F = - m g is the vertical force
therefore
m a = - m g
a = -g constant acceleration down of about 9.81 m/s^2
then
v = 0 - a t = - g t = -9.81 t
h = Hi + 0 t -9.81/2 t^2
h = 1.95 - 4.9 t^2
ground is h = 0
so when it hits the floor
t^2 = 1.95.4.9
solve for t
Now for t seconds, the horizontal velocity was constant = 24.5 m/s
x = v t = 24.5 t
Thank You !!!! I got it correct!
atta go :)
To solve this problem, you can use the basic kinematic equations of motion. The key idea is to break down the motion of the tennis ball into horizontal and vertical components separately.
a) To calculate the horizontal velocity, Vfx, we can use the equation:
Vfx = Vox
Since there is no acceleration in the horizontal direction, the horizontal velocity remains constant throughout the motion. Therefore, the horizontal velocity of the tennis ball when it strikes the court is 24.5 m/s.
b) To find the time, tf, in seconds that the tennis ball moves before striking the court, we need to analyze the vertical motion. We can use one of the kinematic equations:
h = h0 + V0yt - 1/2gt^2
where:
h = height above the tennis court (1.95 m)
h0 = initial vertical position (0 m)
V0y = initial vertical velocity (unknown)
g = acceleration due to gravity (9.8 m/s^2)
t = time (unknown)
Since the ball is dropped (no initial vertical velocity V0y), the equation simplifies to:
h = -1/2gt^2
Substituting the known values:
1.95 = -1/2 * 9.8 * t^2
Simplifying the equation further, we get:
t^2 = 1.95 / (1/2 * 9.8)
Now we can solve for t by taking the square root:
t = sqrt(1.95 / (1/2 * 9.8))
Simplifying this, we find that the time the tennis ball moves before striking the court is approximately 0.442 seconds.
c) To find the horizontal distance xf in meters that the tennis ball moves before striking the court, we can use the equation:
xf = Vx * t
Substituting the known values:
xf = 24.5 m/s * 0.442 s
Simplifying this, we find that the horizontal distance xf is approximately 10.814 meters.
So, to summarize:
a) The tennis ball's horizontal velocity, Vfx, when it strikes the court is 24.5 m/s.
b) The time, tf, in seconds that the tennis ball moves before striking the court is approximately 0.442 seconds.
c) The horizontal distance, xf, in meters that the tennis ball moves before striking the court is approximately 10.814 meters.