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
Answer with Step-by -step explanation:
We are given that
b.
below the positive x-axis
Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=
x-component of vector A=
y-Component of vector A=
Magnitude of vector B=86 m
The vector B makes angle with positive x- axis=
x-component of vector B=
y-Component of vector B=
Vector A=
Vector B=
Vector C=A+B
Substitute the values


c.Direction=
The direction of the vector C=21.5 degree
Answer:
Net force on the wagon is 200 N
Explanation:
As we know by Newton's II law that net force on the system of mass is given as product of mass and acceleration
Here we know that
mass = 100 kg
a = 2 m/s/s
now we have



Answer:
27 m/s
Explanation:
Given:
v₀ = 15 m/s
a = 3 m/s²
t = 4 s
Find: v
v = at + v₀
v = (3 m/s²) (4 s) + (15 m/s)
v = 27 m/s