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
