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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Answer:
a.
b.
Explanation:
We are given that
a.We have to find the angle
b. We have to find the speed
According to law of conservation of momentum
Answer:
t=17.838s
Explanation:
The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:
The acceleration is, by definition:
So, the velocity can be obtained by integrating this expression:
The velocity is, by definition: , so
.
Do x=11 in order to find the time spent.
At this time the velocity is:
This velocity remains constant in the section 2, so for that section the movement equation is:
The left distance is 89 meters, and the velocity is , so:
So, the total time is 14.303+3.5355s=17.838s