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 
If it's a distance graph, then it's a constant speed.
Answer:
1. D
2. D
3. A
The reason why your body goes right and the car car goes left is because your body tries to stay where it was, which is on the right. Your upper body doesn't feel a force and because of that continues in the same direction. Your lower half is pulled out from under you by seat friction to the left, leaving you leaning to the right.
Answer:
5.66 × 10⁻²³ m/s
Explanation:
If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.
Since initial momentum = final momentum,
mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.
My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?
So mv₁ + M × 0 = m × 0 + MV₂
mv₁ = MV₂
V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s
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