Answer:
v = 54 m/s
Explanation:
Given,
The maximum height of the flight of golf ball, h = 150 m
The velocity at height h, u = 0
The velocity of the golf ball right before it hits the ground, v = ?
Using the III equations of motion
<em> v² = u² + 2gh</em>
Substituting the given values in the above equation,
v² = 0 + 2 x 9.8 x 150 m
= 2940
v = 54 m/s
Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s
All the planets revolve around the sun counter clockwise and rotate on hteir axis counterclockwise except venus and neptune.
Hi Randa, want me to help you with physics? The answer is when the object moves in a straight line.
In triangle ABC , using Pythagorean theorem
BC = sqrt(AB² + AC²)
r = sqrt(y² + x²) eq-1
taking derivative both side relative to "t"
dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)
15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)
v₂ = 51.2 m/s
Answer:
Hubble's Discovery
cosmic microwave background radiation
