A ball is kicked with an initial velocity of 21 m/s in the horizontal direction and 16 m/s in the vertical direction. (Assume the ball is kicked from the ground.)
(a) At what speed does the ball hit the ground?
(b) For how long does the ball remain in the air?
(c) What maximum height is attained by the ball?
See previous post: Mon, 9-28-15, 10:47 PM.
(a)33.3 rad/s
(b)500N
(c)40.8m
To find the answers to these questions, we can use the equations of motion for projectile motion. Let's break the problem down step by step.
(a) To find the speed at which the ball hits the ground, we need to determine the horizontal and vertical components of its final velocity. We know that only gravity acts on the vertical component, so it will increase during the motion, while the horizontal component remains constant.
Since the initial vertical velocity is 16 m/s and gravity acts downward at a rate of 9.8 m/s², we can use the formula:
Final vertical velocity (vf) = initial vertical velocity (vi) + (gravity acceleration × time)
0 = 16 - 9.8t
Solving for t gives us:
t = 16 / 9.8
t ≈ 1.63 seconds
Now, let's find the horizontal distance traveled by the ball during this time:
Horizontal distance (d) = horizontal velocity (v₀) × time (t)
d = 21 × 1.63
d ≈ 34.23 meters
The speed at which the ball hits the ground is the same as its final velocity. Using the Pythagorean theorem:
Final velocity (vf) = √(horizontal velocity² + vertical velocity²)
vf = √(21² + 0) = 21 m/s
Therefore, the ball hits the ground with a speed of 21 m/s.
(b) To find out how long the ball remains in the air, we need to calculate the total time it spends in the air. Since the vertical motion is influenced only by gravity, we can calculate the time it takes for the ball to reach its highest point:
At the highest point, the vertical component of velocity becomes zero. Using the formula:
Final vertical velocity (vf) = initial vertical velocity (vi) + (gravity acceleration × time)
0 = 16 - 9.8t
Solving for t:
t = 16 / 9.8
t ≈ 1.63 seconds
Since the ball reaches its highest point in the same amount of time it takes to come back down to the ground, the total time in the air is:
Total time in the air = 2t
Total time in the air ≈ 2 × 1.63
Total time in the air ≈ 3.26 seconds
Therefore, the ball remains in the air for approximately 3.26 seconds.
(c) To find the maximum height attained by the ball, we need to determine the vertical distance traveled during the ascent phase. We can use the formula:
Vertical distance (d) = initial vertical velocity (vi) × time (t) + (1/2) × (gravity acceleration) × (time (t))²
d = 16 × 1.63 + (1/2) × 9.8 × (1.63)²
d ≈ 26.48 meters
Hence, the ball reaches a maximum height of approximately 26.48 meters.