Explanation:
u=166m/s, v=0(at it's highest point final velocity is zero), a=9.8m/s², t=8.6s
by the formula, S=ut+½at².
S=[166×8.6+½.×9.8×(8.6)²]. ...by calculation
S = 1427.6+362.404
S=1790.004m
hope this helps you.
Answer:
The y-component of the electric force on this charge is 
Explanation:
<u>Given:</u>
- Electric field in the region,

- Charge placed into the region,

where,
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

Thus, the y-component of the electric force on this charge is 
Answer: 0.25 m/s
Explanation: Speed = wavelengt · frequency
v = λf and frequency is 1/period f = 1/T
Then v = λ/T = 5 m / 20 s = 0.25 m/s