Answer:
313.6 m downward
Explanation:
The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.
In fact, we have:
where
y(t) is the vertical position of the projectile at time t
h is the initial height of the projectile
is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally
t is the time
a = g = -9.8 m/s^2 is the acceleration due to gravity
We can rewrite the equation as
where the term on the left, 
, represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we find
So the bullet has travelled 313.6 m downward.
Answer:
40,000
Explanation:
200x 200=40000
I'm pretty sure that's the answer
Answer:
v = 14 m/s
= 31.3 mph
The answer would be the same if the mass of the car were 2000 kg
Explanation:
Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.
V^2 = v^2 + 2×a×Δx
Now V, the final velocity is zero as the car stops
0 = v^2 + 2×a×Δx
v^2 = -2×a×Δx
v =√-2×a×Δx .....*
Now applying Newton's Second Law
Fnet = m×a
-Fk = m×a
-μ×N = m×a
-μ×m×g = m×a (The mass cancels out)
a = -μ×g
Substituting the value of a back to equation *
v = √-2×(-μ×g)×Δx
v = √-2×(-0.5×9.8)×20
v = 14 m/s
Therefore the speed the car was travelling with v = 14 m/s
which is equal to 31.3 mph
Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.
