Answer:
Ball hit the tall building 50 m away below 10.20 m its original level
Explanation:
Horizontal speed = 20 cos40 = 15.32 m/s
Horizontal displacement = 50 m
Horizontal acceleration = 0 m/s²
Substituting in s = ut + 0.5at²
50 = 15.32 t + 0.5 x 0 x t²
t = 3.26 s
Now we need to find how much vertical distance ball travels in 3.26 s.
Initial vertical speed = 20 sin40 = 12.86 m/s
Time = 3.26 s
Vertical acceleration = -9.81 m/s²
Substituting in s = ut + 0.5at²
s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²
s = -10.20 m
So ball hit the tall building 50 m away below 10.20 m its original level
The answer is Strontium(Sr)
<span>The inner core is liquid and moving.</span>
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.
Answer:
The magnitudes of the second force is 
The magnitudes of the resultant force is 
Explanation:
From the question we are told that
The force is 
The angle made with second force 
The angle between the resultant force and the first force 
For us to solve problem we are going to assume that
The magnitude of the second force is Z N
The magnitude of the resultant force is R N
According to Sine rule

Substituting values

According to cosine rule

Substituting values

