youre standing on the ground your friend is standing on a balcony of an apartment building 12m above you. you throw a ball straight upward to your friend, it takes 0.536 s to get to her and she catches it when its right in front of her how fast is the ball moving when she catches it

h = Vo*t + 0.5g*t^2 = 12 m.

Vo*0.536 - 4.9*0.536^2 = 12,
Vo = 25.0 m/s.

V = Vo + g*t = 25 - 9.8*0.536 = 19.8 m/s When caught.

To determine the speed of the ball when your friend catches it, we need to calculate its velocity just before it reaches her. Since the ball is thrown straight upward, its velocity will decrease due to the force of gravity.

To solve this, we can use the kinematic equation:

v = u + at

Where:
v = final velocity (unknown)
u = initial velocity (thrown upward, so we can assume it's positive)
a = acceleration (in this case, due to gravity, approximately -9.8 m/s^2)
t = time (0.536 s)

First, we need to find the initial velocity of the ball. We know that the ball is thrown from the ground, so its initial position is at a height of 0 m. The final position is at a height of 12 m because your friend catches it on the balcony.

Using the kinematic equation for position:

s = ut + (1/2)at^2

Where:
s = displacement (12 m)
u = initial velocity (unknown)
a = acceleration (approximated as -9.8 m/s^2)
t = time (0.536 s)

Rearranging the equation, we get:

12 = u(0.536) + (1/2)(-9.8)(0.536)^2

Simplifying the equation:

12 = 0.536u - 1.328

Rearranging again:

0.536u = 12 + 1.328

0.536u = 13.328

Dividing both sides by 0.536:

u ≈ 24.85 m/s

So, the initial velocity of the ball when thrown upward is approximately 24.85 m/s.

Now, let's calculate the final velocity of the ball just before your friend catches it:

v = u + at

v = 24.85 + (-9.8)(0.536)

v ≈ 24.85 - 5.2768

v ≈ 19.5732 m/s

Therefore, the ball is moving at a speed of approximately 19.57 m/s when your friend catches it.