V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}
V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand).... up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)
So v can be calculated.
I’m pretty sure it’s true x
Answer:
The magnitude of the magnetic field is
.
Explanation:
Given that,
Charge, 
Speed of the charged particle, 
The angle between the velocity of the charge and the field is 56°.
The magnitude of force, 
We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

B is the magnetic field.

So, the magnitude of the magnetic field is
. Hence, this is the required solution.
We don't even need to know how many pulses were produced
in those 3 seconds.
The beginning of the first pulse took 3 seconds to travel
45 centimeters from the generator.
Its speed is (45 cm) / (3 sec) = 15 cm/sec.