Answer:
See below
Explanation:
Vertical position = 45 + 20 sin (30) t - 4.9 t^2
when it hits ground this = 0
0 = -4.9t^2 + 20 sin (30 ) t + 45
0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec
max height is at t= - b/2a = 10/9.8 =1.02
use this value of 't' in the equation to calculate max height = 50.1 m
it has 4.22 - 1.02 to free fall = 3.2 seconds free fall
v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL
it will <u>also</u> still have horizontal velocity = 20 cos 30 = 17.32 m/s
total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s
Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m
We can solve the problem by using Newton's second law of motion:

where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object
In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>
B) 1.4 m/s2 horizontally.</span>
I got this answer from the internet
The work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.
<h3>How to calculate work done?</h3>
Work done is a measure of energy expended in moving an object; most commonly, force times distance.
It is said that no work is done if the object does not move, hence, the work done on an object can be calculated as follows:
Work done = Force × Distance
According to this question, a student carries a very heavy backpack and to lift the backpack off the ground, the student must apply 80 N of force to lift the backpack 1.5 m.
Work done = 80N × 1.5m
Work done = 120J
Therefore, the work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.
Learn more about work done at: brainly.com/question/28172139
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