Average velocity is a vector unit (i.e. includes magnitude <em>and </em>direction) calculated by working out distance ÷ time:
80 metres ÷ 20 seconds = 4 metres/seconds (m/s)
Therefore, your final answer is C. 4 m/s south.
Ans. 5 X 10^6
Explanation:
Here,^ represents 6 times 10 and 5 is multiplied. That is 5000 000.
Hope this helps
The equation to be used here is the trajectory of a projectile as written below:
y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity
Since the angle is below horizontal, let's use the minus equation. Substituting the values:
- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m
However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
It is a seismograph that would be most useful.