**Answer:**

A) t = 0.40816 s
, y = 0.916 m

**Explanation:**

A) For this problem we use the kinematic relations

v = v₀ - g t

the highest point zero velocities (v = 0)

t = (v₀-v) / g

t = (4 - 0) / 9.8

t = 0.40816 s

to calculate the height let's use

v² = v₀² - 2 g y

y = vo2 / 2g

y = 4 2 / (2 9.8)

y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

# _photos1 = 0.40916 (30/1)

# _photos1 = 12

yes i take 120 fps

#_fotod = 0.40916 (120/1)

#photos = 5.87 10³

B) The ball is released from a latura h how long it takes to reach the floor

v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

v² = 2gy

v = √ (2 9.8 0.916)

v = √ (2.1397 101)

v = 4.6257 m / s

c) we ask us for the time for latura

y = L / 2

y = 0.916 / 2

y = 0.458 m

now we can use the formula

y = v₀ t - ½ g t²

0.458 = 4.00 t - ½ 9.8 t²

4.9 t² - 4t + 0.458 = 0

t² -0.8163 t +0.09346 = 0

we solve second degree execution

t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

t = [0.8163 ± 0.540] / 2

t₁ = 0.678 m

t₂ = 0.2763 m

the shortest time is for when the ball goes up and the longest when it goes down

D) the graph of vs Vs is expected to be a closed line

and the graph of position versus time a parabola