Answer:
a) net force = 110N to the right.
b) acceleration = 5.5 m/s/s
Explanation:
a) 150N - 40N = 110N
b) F = ma
110N = 20kg x acceleration
acceleration = 110/20 = 5.5
Answer:
a) y = 0.0075 m
b) W = 1.569 J
Explanation:
See attachment for the solution
Answer:
It only depends on the vertical component
Explanation:
Hello!
The horizontal component will tell you how much you travel in that direction.
You could have a large horizontal velocity, but if the vertical velocity is zero, you will never be out of the ground. Similarly, you could have a zero horizontal velocity, but if you have a non-zero vertical velocity you will be some time off the ground. This time can be calculated by two means, one is using the equation of motion (position as a function of time) and the other using the velocity as a fucntion of time.
For the former you must find the time when the position is zero.
Lets consider the origin of teh coordinate system at your feet
y(t) = vt - (1/2)gt^2
We are looking for a time t' for which y(t')=0
0 = vt' - (1/2)gt'^2
vt' = (1/2)gt'^2
The trivial solution is when t'=0 which is the initial position, however we are looking for t'≠0, therefore we can divide teh last equation by t'
v = (1/2)gt'
Solving for t'
t' = (2v/g)
Answer:
1.) Time t = 3.1 seconds
2.) Height h = 46 metres
Explanation:
given that the initial velocity U = 30 m/s
At the top of the trajectory, the final velocity V = 0
Using first equation of motion
V = U - gt
g is negative 9.81m/^2 as the object is going against the gravity.
Substitute all the parameters into the formula
0 = 30 - 9.81t
9.81t = 30
Make t the subject of formula
t = 30/9.81
t = 3.058 seconds
t = 3.1 seconds approximately
Therefore, it will take 3.1 seconds to reach to reach the top of its trajectory.
2.) The height it will go can be calculated by using second equation of motion
h = ut - 1/2gt^2
Substitutes U, g and t into the formula
h = 30(3.1) - 1/2 × 9.8 × 3.1^2
h = 93 - 47.089
h = 45.911 m
It will go 46 metres approximately high.