Explanation:
It is given that,
Speed, v₁ = 7.7 m/s
We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :
So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.
Answer:
huh? do you need help on math?
Explanation:
what do you mean?
Answer:
y = -19.2 sin (23.15t) cm
Explanation:
The spring mass system is an oscillatory movement that is described by the equation
y = yo cos (wt + φ)
Let's look for the terms of this equation the amplitude I
y₀ = 19.2 cm
Angular velocity is
w = √ (k / m)
w = √ (245 / 0.457
w = 23.15 rad / s
The φ phase is determined for the initial condition t = 0 s
, the velocity is negative v (0) = -vo
The speed of the equation is obtained by the derivative with respect to time
v = dy / dt
v = - y₀ w sin (wt + φ)
For t = 0
-vo = -yo w sin φ
The angular and linear velocity are related v = w r
v₀ = w r₀
v₀ = v₀ sinφ
sinφ = 1
φ = sin⁻¹ 1
φ = π / 4 rad
Let's build the equation
y = 19.2 cos (23.15 t + π/ 4)
Let's use the trigonometric ratio π/ 4 = 90º
Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a
y = -19.2 sin (23.15t) cm
Answer:
39.6 m/s
Explanation:
Taking down to be positive:
Given:
v₀ = 0 m/s
a = 9.8 m/s²
Δy = 80 m
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (9.8 m/s²) (80 m)
v = 39.6 m/s
The answer is m/s hope it helps
