Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = (
- v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²
Answer:
Tools that require electricty like drills
Explanation:
Answer:
b) -10 m/s
Explanation:
In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.
Answer:
1) The greatest height attained by the ball equals 20.387 meters.
2) The time it takes for the ball to reach 15 meters approximately equals 1 second.
Explanation:
The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.
thus using third equation of kinematics we obtain the height attained as

where
'v' is the final speed of the ball
'u' is the initial speed of the ball
'a' is the acceleration that the ball is under which in this case equals 9.81 
's' is the distance it covers
Thus for maximum height applying the values in the equation we get

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as
