Answer:
Height as seen by the professor = 38.2 m
Explanation:
Angle of throw = θ = 69°
Velocity of throw = v
X component of velocity = v₁ = v cos 69 = 0.3584 v m/s
Vertical component of the velocity = v₂ = v sin 69 = 0.9336 v m/s
v₂ / v₁ = tan 69 = 2.605
v₂ = 2.605 v₁.
Professor sees as if the x component of velocity =0
v (as seen by professor) + v' = 0
=> v as seen by professor = -v' = -10.5 m/s
This shows that y component of the ball's velocity is 2.605 times its x component of velocity.
with respect to the professor, there is only y component of velocity.
v₂' =v₂ = 2.605 ( -10.5) = 27.4 m/s.
Height as seen by the professor = (27.4)² / 2(9.8) = 38.2 m
Answer:
4m/s²
Explanation:
Initial velocity (u) = 0 m/s
Final velocity (v) = 8 m/s
Time taken (t) = 2 sec
Acceleration (a) = ?
We know

Hope it will help :)
The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.
The Coulomb force between the two balls is:

where

is the Coulomb's constant,

is the intensity of the two charges, and

is the distance between them.
Substituting these numbers into the equation, we get

The force is repulsive, because the charges have same sign and so they repel each other.
1 volt = 1 joule per coulomb
50 volts = 50 joules per coulomb
50 joules/coulomb times 6 coulombs = 300 joules