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?
HELP !
Xo = 21 m/s.
Yo = 16 m/s.
a. V = Vo = Sqrt(Xo^2 + Yo^2).
b. Y = Yo + g*Tr = 0.
16 - 9.8*Tr = 0.
9.8Tr = 16.
Tr = 1.63 s. = Rise time.
Tf = Tr = 1.63 s.
Tr+Tf = 1.63 + 1.63 = 3.26 s. = Time in air.
c. h = 0.5g*Tf^2 = 4.9*1.63^2 = 13 m.
Why did the ball go to the doctor? Because it was feeling a little deflated!
(a) To find the speed at which the ball hits the ground, we just need to find the magnitude of the horizontal velocity. Since the horizontal velocity remains constant throughout the motion, the speed at which the ball hits the ground is also 21 m/s.
(b) To find how long the ball remains in the air, we can use the vertical motion equation: h = v0t + 0.5at^2. Since the initial vertical velocity is 16 m/s and the acceleration due to gravity is -9.8 m/s^2 (assuming upward as positive), we can set h to zero (since the ball hits the ground) and solve for t. So the equation becomes: 0 = 16t + 0.5(-9.8)t^2. Solving this quadratic equation, we find that the ball remains in the air for approximately 3.24 seconds (rounded to two decimal places).
(c) To find the maximum height attained by the ball, we can use the vertical motion equation again. This time, we'll find the time at which the vertical velocity becomes zero, which is the time at the maximum height. Using the equation v = v0 + at, we set the vertical velocity to zero and solve for t: 0 = 16 - 9.8t. Solving this linear equation, we find that the time at the maximum height is approximately 1.63 seconds (rounded to two decimal places). Now we can substitute this time into the vertical motion equation to find the maximum height: h = v0t + 0.5at^2 = 16(1.63) + 0.5(-9.8)(1.63)^2. Solving this equation, we find that the maximum height attained by the ball is approximately 13.15 meters (rounded to two decimal places).
So, to summarize:
(a) The ball hits the ground with a speed of 21 m/s.
(b) The ball remains in the air for approximately 3.24 seconds.
(c) The ball reaches a maximum height of approximately 13.15 meters.
I can help you with that. Let's break down the problem step-by-step.
Step 1: Analyze the vertical motion:
(a) The initial vertical velocity is 16 m/s. Since the ball is kicked from the ground, we can assume the initial height is 0.
Step 2: Calculate the time of flight:
(b) The time it takes for the ball to reach its maximum height can be determined using the formula:
t = (Final velocity - Initial velocity) / acceleration
Since the ball's final vertical velocity is 0 m/s at maximum height, and the acceleration due to gravity is -9.8 m/s^2 (assuming downward direction):
t = (0 - 16) / (-9.8)
Step 3: Calculate the total time of flight:
We know that the time of ascent is the same as the time of descent. Therefore, the total time of flight is twice the time it takes to reach maximum height.
Step 4: Calculate the maximum height:
(c) The maximum height can be determined using the formula:
h = (v^2 - u^2) / (2 * a)
Plugging in the values:
h = (0^2 - 16^2) / (2 * (-9.8))
Step 5: Calculate the horizontal distance traveled:
Since the horizontal velocity is assumed to be constant throughout the motion, we can use the formula:
d = v * t
Now, let's calculate:
Step 1:
(a) The vertical component of the velocity does not affect the speed of the ball hitting the ground. Therefore, the speed at which the ball hits the ground is the same as the horizontal component of the initial velocity.
Speed = 21 m/s
Step 2:
(b) Calculate the time it takes for the ball to reach maximum height:
t = (0 - 16) / (-9.8) = 1.63 seconds
Step 3:
The total time of flight is twice the time it takes to reach maximum height:
Total time = 2 * 1.63 = 3.26 seconds
Step 4:
(c) Calculate the maximum height:
h = (0^2 - 16^2) / (2 * (-9.8)) = 12.65 meters
Step 5:
Calculate the horizontal distance traveled:
Since the horizontal velocity is constant throughout the motion, the horizontal distance traveled can be calculated using:
d = v * t = 21 m/s * 3.26 s = 68.46 meters
So, to summarize:
(a) The ball hits the ground at a speed of 21 m/s.
(b) The ball remains in the air for 3.26 seconds.
(c) The maximum height attained by the ball is 12.65 meters.
(d) The horizontal distance traveled by the ball is 68.46 meters.
Sure, I can help you with that!
To find the answer to each question, we can use the equations of motion for projectile motion. These equations are:
1) Vertical displacement, y = ut + (1/2)gt^2
2) Final velocity in the vertical direction, vy = u + gt
3) Final velocity in the horizontal direction, vx = u
(a) To find the speed at which the ball hits the ground, we need to find the magnitude of the final velocity vector. The final velocity vector is the resultant of the horizontal and vertical components of velocity.
Since the ball is kicked horizontally, there is no change in its horizontal velocity. So, the horizontal component of the velocity remains 21 m/s.
In the vertical direction, the initial velocity is 16 m/s, and the acceleration due to gravity is approximately 9.8 m/s^2 downwards. The final velocity in the vertical direction can be found using equation (2):
vy = u + gt
vy = 16 m/s + (9.8 m/s^2)(t)
At the moment the ball hits the ground, the vertical displacement is zero. So, we can find the time it takes for the ball to hit the ground by solving for t in equation (1), with y = 0 and u = 16 m/s:
0 = (16 m/s)t + (1/2)(9.8 m/s^2)(t^2)
0 = 16t + 4.9t^2
Solving this quadratic equation will give us the time it takes for the ball to hit the ground.
Once we have the time, we can substitute it into equation (2) to find the final velocity in the vertical direction, vy. Finally, we can use the Pythagorean theorem to find the speed at which the ball hits the ground by taking the square root of the sum of the squares of the horizontal and vertical components of velocity.
(b) To find how long the ball remains in the air, we will use the same equation as in part (a), y = ut + (1/2)gt^2, but this time we solve for the time it takes for the ball to reach its highest point, where y = maximum height.
(c) To find the maximum height attained by the ball, we can find the vertical displacement at the point where the ball reaches its highest point.
I hope this explanation helps you understand how to find the answers to these questions! Let me know if you need further assistance with the calculations.