Answer:
A. Vx = 3.63 m/s
B. Vy = -45.73 m/s
C. |V| = 45.87 m/s
D. θ = -85.46°
Explanation:
Given that position, r, is given as:
r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk
Velocity is the derivative of position, r:
V = dr/dt = 3.63 - 11.46t^j
A. x component of velocity, Vx = 3.63 m/s
B. y component of velocity, Vy = -11.46t
t = 3.99 secs,
Vy = - 11.46 * 3.99 = -45.73 m/s
C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]
|V| = √(2091.2329 + 13.1769)
|V| = √(2104.4098)
|V| = 45.87 m/s
D. Angle of the velocity relative to the x axis, θ is given as:
tanθ = Vy/Vx
tanθ = -45.73/3.63
tanθ = -12.6
θ = -85.46°
Answer:
the answer is a. a ball is moving towards the camera faster then slower
Answer:
K =6.697 Kg/s²
Explanation:
Given:
delta m =41 g = 0.041 kg
delta x = 6cm = 0.06m
g = 9.8 m/s²
according to the given formula
K = delta m g /delta x
K = (0.041 kg × 9.8 m/s²) / 0.06m
K =6.697 Kg/s²
Answer:
Option (b) is correct.
Explanation:
The motion under the influence of gravity is called projectile motion.
The acceleration due to gravity is constant through out the motion and it is always acting downwards.
When an athlete jumps and follow the projectile path, it always have the same horizontal velocity as there is no acceleration in the horizontal direction.
Also he has the vertical acceleration constant which is equal to the acceleration due to gravity and acts towards the center of earth.
Option (b) is correct.