A ball is thrown vertically upward with a speed of +12.0 m/s.
How long does the ball take to hit the ground after it reaches its highest point?
What is its velocity when it returns to the level from which it started?
Well, the easy way is to say the average speed of the ball on the awy up is (12 + 0) / 2 = 6 m/s
Now how high did it go?
(1/2) m v^2 = m g h
144/2 = 9.8 h
h = 7.35 meters up
so
6 t = 7.35
t = 1.22 seconds
part 2 is trivial, same speed down as up but opposite direction, -12 m/s
I do not like the way you did this questrion
To find the time it takes for the ball to hit the ground after reaching its highest point, we can use the fact that the time taken to reach the highest point is equal to the time taken to return back to the original level.
Let's calculate the time it takes for the ball to reach its highest point first.
Given:
Initial velocity (u) = +12.0 m/s (positive since the ball is thrown upward)
We know that the final velocity (v) at the highest point will be zero, as the ball momentarily stops before falling back down.
Using the equation v = u + at, where a is the acceleration due to gravity (-9.8 m/s²), and v is 0, we can solve for time (t):
0 = 12.0 m/s - 9.8 m/s² * t
Rearranging the equation to solve for t:
9.8 m/s² * t = 12 m/s
t = 12 m/s / 9.8 m/s²
t ≈ 1.22 seconds
Therefore, it takes approximately 1.22 seconds for the ball to reach its highest point.
Next, since the time taken to reach the highest point is equal to the time taken to return back to the original level, the ball will take approximately 1.22 seconds to hit the ground.
Now, let's find the velocity of the ball when it returns to the level from which it started.
Since the velocity of the ball when it is thrown upward is +12.0 m/s, the velocity when it returns at the same level will be negative.
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (-9.8 m/s²), and t is the time taken, we can find the final velocity (v):
v = 12.0 m/s + (-9.8 m/s²) * 1.22 s
v ≈ -12.8 m/s
Therefore, the velocity of the ball when it returns to the level from which it started is approximately -12.8 m/s.
To find the time it takes for the ball to hit the ground after reaching its highest point, we need to use the equations of motion.
First, we need to identify the initial velocity (u), the final velocity (v), the acceleration (a), and the time taken (t). In this case, the ball is thrown vertically upward, so the acceleration due to gravity (a) is -9.8 m/s^2 (negative because it acts in the opposite direction of the initial velocity). The initial velocity (u) is +12.0 m/s (positive because it is directed upward), and the final velocity (v) is 0 m/s (since the ball comes to a stop at its highest point).
We can use the equation v = u + at to find the time it took for the ball to reach its highest point:
0 = 12.0 + (-9.8)t
Simplifying the equation:
-12.0 = -9.8t
Dividing both sides by -9.8, we get:
t = 12.0 / 9.8 ≈ 1.22 seconds
Therefore, the ball takes approximately 1.22 seconds to hit the ground after reaching its highest point.
To find the velocity when the ball returns to the level it started, we can use the same equation v = u + at, but with the time taken as twice the time it took to reach the highest point (since the upward and downward journey take the same amount of time).
t_total = 2 * t = 2 * 1.22 ≈ 2.44 seconds
Using this value for t_total, we can now find the final velocity (v).
v = u + at
v = 12.0 + (-9.8) * 2.44
v ≈ -11.2 m/s
Therefore, when the ball returns to the level from which it started, its velocity is approximately -11.2 m/s (negative because it is directed downward).