Given that,
Distance in south-west direction = 250 km
Projected angle to east = 60°
East component = ?
since,
cos ∅ = base/hypotenuse
base= hyp * cos ∅
East component = 250 * cos 60°
East component = 125 km
<u>Answer:</u>
Cannonball will be in flight before it hits the ground for 2.02 seconds
<u>Explanation:</u>
Initial height from ground = 20 meter.
We have equation of motion , 
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 
, we need to calculate time when s = 20 meter.
Substituting
So it will take 2.02 seconds to reach ground.
Answer:
D) 42.87 m/s
Explanation:
First, find the time it takes him to land. Given in the y direction:
Δy = 60 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
60 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 3.5 s
Next, find the speed needed to travel the horizontal distance in that time. Given in the x direction:
Δx = 60 m
a = 0 m/s²
t = 3.5 s
Find: v₀
Δy = v₀ t + ½ at²
150 m = v₀ (3.5 s) + ½ (0 m/s²) (3.5 s)²
v₀ = 42.87 m/s
It is not only the horizontal part of the loop that dips
into the magnetic field. We can neglect or disregard the horizontal parts of
the loop that hollows into the magnetic fields since only the parts
perpendicular to the magnetic field complement to it.
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