Given that,
Initial velocity , Vi = 0
Final velocity , Vf = 40 m/s
Acceleration due to gravity , a = 9.81 m/s²
Distance can be calculated as,
2as = Vf² - Vi²
2 * 9.81 *s = 40² - 0²
s = 81.55 m
For half height, that is, s = 40.77m
Vf= ??
2as = Vf² - Vi²
2 * 9.81 * 40.77 = Vf² - 0²
Vf² = 800
Vf = 28.28 m/s
Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.
<h2>
Time taken is 0.459 seconds</h2>
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = 81 km/hr = 22.5 m/s
Time, t = ?
Acceleration, a = 49 m/s²
Substituting
v = u + at
22.5 = 0 + 49 x t
t = 0.459 seconds
Time taken is 0.459 seconds
Answer:
(C) 40m/s
Explanation:
Given;
spring constant of the catapult, k = 10,000 N/m
compression of the spring, x = 0.5 m
mass of the launched object, m = 1.56 kg
Apply the principle of conservation of energy;
Elastic potential energy of the catapult = kinetic energy of the target launched.
¹/₂kx² = ¹/₂mv²
where;
v is the target's velocity as it leaves the catapult
kx² = mv²
v² = kx² / m
v² = (10000 x 0.5²) / (1.56)
v² = 1602.56
v = √1602.56
v = 40.03 m/s
v ≅ 40 m/s
Therefore, the target's velocity as it leaves the spring is 40 m/s
Answer:
Velocity,v = 0.323 m/s
Explanation:
The acceleration of a particle is given by :

b = 0.8 m when x = 0
Since,

![\dfrac{v^2}{2}=-[0.1x-0.8cos\dfrac{x}{0.8}]+c](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%5E2%7D%7B2%7D%3D-%5B0.1x-0.8cos%5Cdfrac%7Bx%7D%7B0.8%7D%5D%2Bc)
At x = 0, v = 1 m/s


![\dfrac{v^2}{2}=-[0.1x-0.8cos\dfrac{x}{0.8}]-0.3](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%5E2%7D%7B2%7D%3D-%5B0.1x-0.8cos%5Cdfrac%7Bx%7D%7B0.8%7D%5D-0.3)
At x = -1 m


v = 0.323 m/s
So, the velocity of the particle is 0.323 m/s. Hence, this is the required solution.
I think its a higher frequency