It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.
From the question given above, the following data were obtained:
Height (H) = 206 m
<h3>Time (t) =? </h3>
NOTE: Acceleration due to gravity (g) = 10 m/s²
The time taken for the ball to get to the ground can be obtained as follow:
H = ½gt²
206 = ½ × 10 × t²
206 = 5 × t²
Divide both side by 5
Take the square root of both side
<h3>t = 6.42 s</h3>
Therefore, it will take 6.42 s for the ball to get to the ground.
Learn more: brainly.com/question/24903556
Answer:
25 m/s
Explanation:
Centripetal acceleration is the square of the tangential velocity divided by the radius.
a = v² / r
15.625 m/s² = v² / (40 m)
v² = 625 m²/s²
v = 25 m/s
The speed of the car is 25 m/s.
Answer:
a) Acceleration is zero
, c) Speed is cero
Explanation:
a) the equation that governs the simple harmonic motion is
x = A cos (wt +φφ)
Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition
Body acceleration is
a = d²x / dt²
Let's look for the derivatives
dx / dt = - A w sin (wt + φ)
a = d²x / dt² = - A w² cos (wt + φ)
In the instant when it is not stretched x = 0
As the spring is released at maximum elongation, φ = 0
0 = A cos wt
Cos wt = 0 wt = π / 2
Acceleration is valid for this angle
a = -A w² cos π/2 = 0
Acceleration is zero
b)
c) When the spring is compressed x = A
Speed is
v = dx / dt
v = - A w sin wt
We look for time
A = A cos wt
cos wt = 1 wt = 0, π
For this time the speedy vouchers
v = -A w sin 0 = 0
Speed is cero
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