Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
By equation of motion we have v = u + at
Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration
Here v = 141 m/s, u = 17.7 m/s and t = 6 s
On substitution we will get
141 = 17.7+ 6a
So, a = (141-17.7)/6 = 20. 55 m/
Aceeleration = 20. 55 m/ along north direction.
Answer:
they cant travel through a vacuum
Acceleration = (0.2 x g) = 1.96m/sec^2.
<span>Accelerating force on 1kg. = (ma) = 1.96N. </span>
<span>1kg. has a weight (normal force) of 9.8N. </span>
<span>Coefficient µ = 1.96/9.8 = 0.2 minimum. </span>
<span>Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. </span>
<span>e.g. 75kg = (75 x g) = 735N. </span>
<span>Accelerating force = (735 x 0.2) = 147N</span>