The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.
The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.
v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s
Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.
The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.
Combining this together we get:
(1) vx=40m/s and vy=10m/s
B. is the answer.
C is not correct because the light is actually reflected off of an opaque object.
Answer:
1). Average speed = 1.5 m per second
2). Average velocity = 1.5 m per second
Explanation:
1). Since, speed is a scalar quantity
Therefore, average speed of the trip = 
From the graph attached,
Total distance covered = 10 + 10 + 20 + 0 + 20 + 30
= 90 meters
Total time taken = 60 seconds
Average speed = 
= 1.5 meter per second
2). Velocity is a vector quantity.
Therefore, average velocity = 
= 
= 
= 1.5 meter per second
Answer:
L = - 1361.591 k Kgm/s
Explanation:
Given
mA = 55.2 Kg
vA = 3.45 m/s
rA = 6.00 m
mB = 62.4 Kg
vB = 4.23 m/s
rB = 3.00 m
mC = 72.1 Kg
vC = 4.75 m/s
rC = - 5.00 m
then we apply the equation
L = (mv x r)
⇒ LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s
⇒ LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s
⇒ LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s
Finally, the total counterclockwise angular momentum of the three joggers about the origin is
L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k Kgm/s
L = - 1361.591 k Kgm/s
