Initially, the velocity vector is 
. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 
, so the velocity is 
.
Converting back to direction and magnitude, we get 
Answer:
D = 9.9 10⁶ mi
Explanation:
In the exercise they give the expression for maximum viewing distance
D = 2 r h + h²
Ask for this distance for a height of 1100 feet
Let's calculate
D = 2 3960 1100 + 1100²
D = 8.712 10⁶ + 1.21 10⁶
D = 9.92 10⁶ mi
D = 9.9 10⁶ mi
Answer:
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Answer:
W = 145.8 [N]
Explanation:
To solve this problem we must remember that weight is defined as the product of mass by gravity, in this case lunar gravity.
W = m*g
where:
m = mass = 90 [kg]
g = gravity acceleration = 1.62 [kg/m²]
W = 90*1.62
W = 145.8 [N]
To answer this question, first we take note that the maximum height that can be reached by an object thrown straight up at a certain speed is calculated through the equation,
Hmax = v²sin²θ/2g
where v is the velocity, θ is the angle (in this case, 90°) and g is the gravitational constant. Since all are known except for v, we can then solve for v whichi s the initial velocity of the projectile.
Once we have the value of v, we multiply this by the total time traveled by the projectile to solve for the value of the range (that is the total horizontal distance).
