Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
know v0 and R, determine the general expressions for the two distinct launch angles θ1 and θ2 that will allow the projectile to hit D. For v0 = 42 m/s and R = 70 m, determine numerical values for θ1 and θ2?
1 answer:
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.
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Explanation:
It is given that,
Total weight of the piston, W = F = 70 N
Area of the piston,
Let P is the pressure exerted on the piston by the gas. The force per unit area is called the pressure exerted pressure of the gas. Mathematically, it is given by :
We know that the atmospheric pressure is given by :
So, the pressure is given by :
Hence, this is the required solution.
Answer:
a. 7.046 Nm²/C
b. 2.348 Nm²/C
Explanation:
Data given:
Base of equilateral triangle = 25.0 cm = 0.25 m
Strength of electric field = 260 N/C
In order to find the electric flux we first have to find out the area of triangle.
Area of triangle =
=
= 0.0271 m³
Lets find electric flux,
Electric Flux = E. A
= 260×0.0271
= 7.046 Nm²/C
Now we can find the electric flux through each of the three sides.
Electric flux through three sides =
= 2.348 N m²/C