Answer:
Approximately (assuming that the projectile was launched at angle of above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing is the same as the altitude at which this projectile was launched: .
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is (upwards,) the vertical velocity right before landing would be (downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be . In other words, .
Hence, the time it takes to achieve a (vertical) velocity change of would be:
.
Hence, this projectile would be in the air for approximately .
You should just ask the wave
Answer:
Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.
Explanation:
Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass,
Area of cross section,
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
R = W
or
Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,
So, the drag coefficient is 1.01.