Answer:
A) t = 0.40816 s, y = 0.916 m
Explanation:
A) For solving this problem, we utilize kinematic equations.
v = v₀ - g t
For the highest point, the velocity is zero (v = 0).
t = (v₀ - v) / g
t = (4 - 0) / 9.8
t = 0.40816 s
To calculate the height, we apply the equation
v² = v₀² - 2 g y
y = v₀² / (2g)
y = 4² / (2 * 9.8)
y = 0.916 m
To ascertain the number of photos, we can apply a direct proportionality method: if you take 30 photos in a second for 0.40816 s, how many photos are taken?
# _photos1 = 0.40816 (30/1)
# _photos1 = 12
Yes, I capture at 120 fps.
#_fotod = 0.40816 (120/1)
#photos = 5.87 * 10³
B) The ball is dropped from height h; we want to find the time it takes to reach the ground.
v² = v₀² + 2 g y
where the initial velocity is zero, and the velocity the expert releases is equal to the outgoing speed v = 4 m/s.
v² = 2gy
v = √(2 * 9.8 * 0.916)
v = √(2.1397 * 101)
v = 4.6257 m/s
C) We need to calculate the two instances when the ball reaches a height of y = ½ h. The final formulation and numerical outcomes will be provided.
Now applying the formula
y = L / 2
y = 0.916 / 2
y = 0.458 m
At this point, we can apply the equation
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 need to solve the quadratic equation.
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 period is when the ball ascends, while the longest is during its descent.
D) The expected plot of vs versus time is anticipated to form a closed loop, while the position against time should resemble a parabola.
