Johnny threw a ball upward with an initial velocity of 2000 cm /s and was able to catch it before it reached the ground on its return.
a. What is the velocity of the ball when it reaches its maximum hieght?
b. What was its final velocity just before it touched Johnny's hand?
c. What was its final velocity when Johnny catches it?
if straight up then zero at the top
-2000 cm/s = -20 m/s as it hits his glove
zero if his hand stops it.
a. Well, when the ball reaches its maximum height, its velocity will be zero. Just like my chances of catching a Frisbee – zero!
b. Right before the ball touches Johnny's hand, its final velocity will also be zero. Kind of like my motivation on a Monday morning – zero!
c. When Johnny catches the ball, its final velocity will be the same as the initial velocity he threw it with – 2000 cm/s. Just like my excitement when I see a plate of cookies – 2000%!
Remember, it's just physics – no clowning around!
To solve the questions, we can use the equations of motion for an object in free fall. Here are the step-by-step calculations:
a. To find the velocity of the ball when it reaches its maximum height, we can use the formula:
Final Velocity (vf) = Initial Velocity (vi) + Acceleration (a) * Time (t)
At the highest point, the ball momentarily comes to rest, so its final velocity is 0 cm/s. The acceleration, in this case, is due to gravity and is approximately -980 cm/s^2 (because it acts in the opposite direction to the initial velocity). The initial velocity is given as 2000 cm/s.
0 = 2000 + (-980) * t
Solving for t:
-980t = -2000
t = -2000 / -980
t ≈ 2.04 seconds
Now, let's substitute t back into the equation to find the velocity at the maximum height:
vf = 2000 + (-980) * 2.04
vf ≈ -998.32 cm/s
Therefore, the velocity of the ball when it reaches its maximum height is approximately -998.32 cm/s.
b. To find the final velocity just before it touches Johnny's hand, we can use the same equation:
vf = vi + a * t
At the highest point, the velocity is -998.32 cm/s. The acceleration remains the same (-980 cm/s^2) as it acts downward. The time can be calculated by doubling the previous value (since the upward and downward paths are symmetrical).
vf = 2000 + (-980) * 2 * t
-998.32 = 2000 + (-980) * 2 * t
Solving for t:
-1960t = -2998.32
t ≈ 1.53 seconds
Now, substitute t back into the equation to find the final velocity just before Johnny catches it:
vf = 2000 + (-980) * 2 * 1.53
vf ≈ 65.68 cm/s
Therefore, the final velocity just before it touches Johnny's hand is approximately 65.68 cm/s.
c. Finally, to find the final velocity when Johnny catches the ball, we need to consider that the initial velocity on the downward path is -2000 cm/s.
Using the equation:
vf = vi + a * t
vf = -2000 + (-980) * t
Substitute the value of t we found in part b:
vf = -2000 + (-980) * 1.53
vf ≈ -2820.4 cm/s
Therefore, the final velocity when Johnny catches the ball is approximately -2820.4 cm/s.
To answer these questions, we need to understand the basic principles of projectile motion.
a. What is the velocity of the ball when it reaches its maximum height?
When a ball is thrown upward, its velocity decreases due to the force of gravity. At its maximum height, the velocity becomes zero before it starts coming down. To find the velocity at the maximum height, we need to divide the initial velocity by 2. This is because the time it takes for the ball to reach its maximum height is half of the total time of flight.
To determine the velocity at the maximum height, we can use the equation:
v = u - gt
Where:
v = final velocity (which is zero at the maximum height)
u = initial velocity (2000 cm/s)
g = acceleration due to gravity (approximately 980 cm/s^2)
t = time taken to reach the maximum height (which is half of the total time of flight)
Given that:
u = 2000 cm/s
g = 980 cm/s^2
We can rearrange the equation to solve for t:
t = (u - v) / g
Substituting the values, we have:
t = (2000 cm/s - 0 cm/s) / 980 cm/s^2
t ≈ 2.04 seconds
Now that we have the time taken to reach the maximum height, we can find the velocity at that point:
v = u - gt
v = 2000 cm/s - (980 cm/s^2)(2.04 s)
v ≈ 1978.4 cm/s
Therefore, the velocity of the ball when it reaches its maximum height is approximately 1978.4 cm/s.
b. What was its final velocity just before it touched Johnny's hand?
To find the final velocity just before the ball touches Johnny's hand, we can use the same equation:
v = u - gt
Given that:
u = 2000 cm/s (initial velocity)
g = 980 cm/s^2 (acceleration due to gravity)
t = time taken to reach the ground
We don't have the exact value of t, but we can calculate it by using another equation:
h = ut - (1/2)gt^2
Where:
h = height (in this case, the total height from the initial point to the ground)
u = initial velocity (2000 cm/s)
g = acceleration due to gravity (980 cm/s^2)
t = total time of flight
In this case, the total height is 0 because the ball ends up at the starting height (Johnny's hand) again. Therefore, we can set h = 0 and solve for t:
0 = (2000 cm/s)(t) - (1/2)(980 cm/s^2)(t^2)
Simplifying the equation gives us a quadratic equation:
-490t^2 + 2000t = 0
Factoring out t, we have:
t(2000 - 490t) = 0
This gives us two solutions: t = 0 seconds (which is not possible) and t = 2000/490 ≈ 4.08 seconds.
Now that we have the time taken to reach the ground, we can find the final velocity just before touching Johnny's hand:
v = u - gt
v = 2000 cm/s - (980 cm/s^2)(4.08 s)
v ≈ -2072.8 cm/s
Note that the negative sign indicates that the ball has a downward velocity.
Therefore, the final velocity just before the ball touches Johnny's hand is approximately -2072.8 cm/s.
c. What was its final velocity when Johnny catches it?
The final velocity when Johnny catches the ball is the same as the velocity just before it touches his hand. Therefore, the final velocity when Johnny catches the ball is approximately -2072.8 cm/s.