A ball is thrown straight up at an initial speed of 24 m/s. How long did it take to reach maximum height? How high did it rise before falling again?
I am confused on how to attempt this problem. Please provide me with some guidance:)
v = Vi - g t
0 = 24 - 9.81 t
t = 24/9.81
h = 24 t -4.9 t^2
I get it now! I was confused about the final velocity. So, is the final velocity 0 m/s because that is when it hit the ground?
No, 0 at max height. It stops at the top, then falls
Oh...So it pauses for a second and then falls to the ground!
Do you mind if I verify a couple more questions? My physics background is not very strong:(
go ahead and post them separately above.
Thanks!
well, it pauses for in ten zillionth of a second. In other words the velocity goes from upward to downward so it goes through zero :)
To find the time it takes for the ball to reach the maximum height, we can use the fact that the vertical velocity of the ball at that point is zero. This is because at the highest point of the ball's trajectory, it momentarily stops moving upward before reversing direction and falling back downwards.
We know that the initial upward velocity of the ball is 24 m/s, and the acceleration due to gravity is approximately -9.8 m/s^2 (assuming the ball is thrown on Earth). With this information, we can use the following equation to find the time it takes to reach the maximum height:
v = u + at
Here, v represents the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the ball stops moving at its highest point, v = 0. So, we have:
0 = 24 + (-9.8)t
Simplifying this equation, we get:
-24 = -9.8t
Dividing both sides by -9.8, we find:
t = -24 / -9.8
t ≈ 2.45 seconds
Therefore, it takes approximately 2.45 seconds for the ball to reach maximum height.
To find the maximum height the ball rises before falling again, we can use the kinematic equation:
s = ut + (1/2)at^2
Where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
At the highest point, the final velocity is 0, and we are interested in finding the displacement. So, we have:
0 = 24t + (1/2)(-9.8)t^2
Simplifying this equation, we get:
0 = 24t - 4.9t^2
Rearranging and factoring out t, we have:
4.9t^2 - 24t = 0
Dividing both sides by t, we get:
4.9t - 24 = 0
From this equation, we can find the two possible values of t. However, since we are interested in the time it takes to reach the maximum height, we neglect the solution t = 0 (as it corresponds to the initial position of the ball).
Solving for t, we find:
t ≈ 4.9 / 24
t ≈ 0.204 seconds
Substituting this value of t back into the displacement equation, we get:
s = 24(0.204) + (1/2)(-9.8)(0.204)^2
s ≈ 4.94 meters
Therefore, the ball rises approximately 4.94 meters before falling down again.