A tennis player hits a ball at ground level, giving it an initial velocity of 24 m/s at 57 degrees above the horizontal. (a) what are the horizontal and verticla components of the ball;s initial velocity? (b) how high above the ground does the ball go? (c) How long does it take the ball to reach its maximum height? (d) What are the ball;s velocity and acceleration at its highest point? (e) for how long a time is the ball in the air? (f) When this ball lands on the court, how far is it from the place where it was hit?
u = initial velocity = 24 m/s
θ = angle = 57 degrees
A. components
horizontal component, uh
= u cos(θ)
= 24*cos(57)
= 13.07 m/s
vertical component, uv
= u sin(θ)
= 24*sin(57)
= 20.13 m/s
B. Max height, S
consider vertical direction only
uv = 20.13 (initial)
vv = 0 (final, at highest point)
a= -g = -9.81 m/s/s
2*a*S = vv2 - uv2
2*(-9.81)*S = -20.132
S=20.65 m
C. Time to reach max. height, t
Consider vertical direction only:
v=u+at
0 = 20.13 + (-9.81)*t
t = 20.13/9.81 = 2.05 s.
D. At highest point,
velocity =sqrt(02+uh2)
=uh
=13.07 m/s (horizontal)
acceleration
= -g
= -9.81 m/s/s
E. Ball in the air, T
Time = 2t = 2*2.05 s = 4.1 s.
F. Horizontal distance travelled, Sh
=uh * T
= 13.07 m/s * 4.1 s
= 53.6 m.
The solving is so cute and great with proper explanation
Thanks
!
(a) Well, the initial velocity of the ball has both horizontal and vertical components. To find the horizontal component, we can use some trigonometry. So, let's see, *puts on math clown wig* the horizontal component is given by multiplying the initial velocity (24 m/s) by the cosine of the angle (57 degrees). Therefore, the horizontal component is 24 m/s * cos(57 degrees).
As for the vertical component, it can be found similarly. It is equal to the initial velocity (24 m/s) multiplied by the sine of the angle (57 degrees). Easy peasy lemon squeezy!
(b) To find out how high the ball goes, we need to use the vertical component of the initial velocity. Now, we can delve into some physics clownery. We can use the following kinematic equation: height = (initial velocity squared * sine squared of the angle) divided by (2 times the acceleration due to gravity).
(c) The time it takes for the ball to reach its maximum height is the time when the vertical component of its velocity becomes zero. So, it reaches its peak, and I'm not talking about mountains here. We can calculate this using the following equation: time = (final velocity - initial velocity) divided by (acceleration due to gravity).
(d) At its highest point, the ball's velocity is...drum roll please...zero! Yep, you heard it right, it momentarily stops. As for the acceleration, it stays the same as the acceleration due to gravity, making it feel down and sad...or maybe just fall effect.
(e) The time the ball spends in the air can be determined by finding the time it takes for the ball to go from its highest point to the ground. Remember, what goes up must come down! So, we use the same equation as before: time = (final velocity - initial velocity) divided by (acceleration due to gravity).
(f) Finally, to find out how far the ball is from the spot where it was hit, we can use the horizontal component of the initial velocity and the time the ball is in the air. So, distance = horizontal component of initial velocity multiplied by time.
I hope that clears things up, or maybe just adds a bit more clown confusion. Enjoy your tennis physics adventure!
To answer these questions, we will break down the initial velocity into its horizontal and vertical components. Using the given information:
Initial velocity (v0) = 24 m/s
Launch angle (θ) = 57 degrees
(a) The horizontal component of the initial velocity can be found using the equation:
v0x = v0 * cos(θ)
Substituting the values:
v0x = 24 m/s * cos(57 degrees)
v0x ≈ 13.02 m/s
Therefore, the horizontal component of the initial velocity is approximately 13.02 m/s.
The vertical component of the initial velocity can be found using the equation:
v0y = v0 * sin(θ)
Substituting the values:
v0y = 24 m/s * sin(57 degrees)
v0y ≈ 19.02 m/s
Therefore, the vertical component of the initial velocity is approximately 19.02 m/s.
(b) To determine the maximum height above the ground (h), we need to use the vertical component of the initial velocity and the acceleration due to gravity (g).
We know that vertical displacement (h) can be calculated using the formula:
h = (v0y^2) / (2g)
Substituting the values:
h = (19.02 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 19.55 m
Therefore, the ball reaches a height of approximately 19.55 meters above the ground.
(c) The time it takes for the ball to reach its maximum height can be calculated using the formula:
t = v0y / g
Substituting the values:
t = 19.02 m/s / 9.8 m/s^2
t ≈ 1.94 s
Therefore, it takes approximately 1.94 seconds for the ball to reach its maximum height.
(d) At its highest point, the ball's vertical velocity (vy) is zero since it momentarily stops. The horizontal velocity (vx) is constant throughout the projectile motion. The acceleration (a) is equal to the acceleration due to gravity (g).
Therefore, the ball's velocity at its highest point is (vx, vy) = (13.02 m/s, 0 m/s), and the acceleration is (0 m/s^2, -9.8 m/s^2) or simply (0, -9.8) m/s^2.
(e) The total time of flight, or the time the ball is in the air, can be calculated by doubling the time it took to reach the maximum height since the ball follows a symmetrical trajectory.
Total time of flight = 2 * t ≈ 2 * 1.94 s
Total time of flight ≈ 3.88 s
Therefore, the ball is in the air for approximately 3.88 seconds.
(f) To find the horizontal distance traveled by the ball when it lands, we need to calculate the horizontal displacement. Assuming the landing point is the same as the starting point, the horizontal displacement is given by the formula:
d = v0x * t
Substituting the values:
d = 13.02 m/s * 3.88 s
d ≈ 50.50 m
Therefore, when the ball lands, it is approximately 50.50 meters away from the point where it was hit.