The bullet travels a horizontal distance of 276.5 m
The bullet is shot forward with a horizontal velocity 
. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>
The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.
The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.
Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,
Substitute 0 m/s for 
, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.
The horizontal distance traveled by the bullet is given by,
Substitute 500 m/s for and 0.5530s for t.
The bullet travels a distance of 276.5 m.
Answer:
we encounter ocassion where one or more object move in the which is not stationary with respect to another example a boat is cross a river that is flowing at some rate of aeroplane encountring wind durning it motion
Answer:
Hello your question is missing some parts attached below is the missing part of your question
answer: <em>many primary sensory Neurons will converge and become a single Neuron and the single Neuron will send a single harmonized signal to the Brain</em>.
Explanation:
The reason regardless of the location that will make you perceive the two points as a single point rather than as two distinct points is that many primary sensory Neurons will converge and become a single Neuron and the single Neuron will send a single harmonized signal to the Brain.
Answer:
The formula for force according to Newton's second law of motion is F=ma or force for an object to move is equal to mass times acceleration.
Acceleration or average acceleration defined as change in velocity per time.
F=ma
F=1.2x10³kg*(20m/s)(1/5s)=4.8x10³ Newtons
Explanation:
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Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
