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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Given that,
The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.
(a) Capacitance of capacitor 1,
Capacitance of capacitor 2,
Capacitance of capacitor 3,
(b) The equivalent capacitance in series combination is :
Hence, this is the required solution.