Projectile in Motion: horizontal

AI Thread Summary
The discussion centers on solving a projectile motion problem involving a ball fired at an initial speed of 1.70 x 10^3 m/s and an angle of 55 degrees. Part A requires calculating the horizontal range, while Part B asks for the total time of flight. The problem is approached using kinematic formulas, emphasizing the separation of horizontal and vertical velocity components. Key formulas include the time of flight, t = (2Vsin θ)/g, and the horizontal distance, d_H = (V^2 sin 2θ)/g. Participants are encouraged to substitute the given values into these formulas to find the solutions.
khai06
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i'm having trouble with this problem, so i want to share it with you guys so some of you can briefly describe what is it asking for and how to solve using what fomula... the question is

A ball is fired from the ground with an initial speed of 1.70 x 10^3 m/s (which is approximately five times the speed of sound) at an initial angle of 55.0 degree to the horizontal. Neglecting air resistance, find the following

A. the ball's horizonal range
B. the amount of time the ball is in motion...

initial Velocity = 1.70 x 10^3
initial angle = 55.0 degree

what formula should i use?
 
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What is it asking for?

Part A is asking how far the projectile moves horizontally from its initial position

Part B is self explanatory.

Think: what do you know? You know that, by definition, a projectile is a flying object that, once given an initial velocity, undergoes motion governed solely by the force of gravity. Hence, this is a constant acceleration problem. Which kinematics formulas should you apply to such a problem?
 
There are two velocity components : horizontal velocity ( 1.70 x 10^3 x cos55 ) that remais the same, and vertical velocity (1.70 x 10^3 x sin55 ) that is modified by the gravity acceleration along the flight.
So...
 
here is the derrivation of the formulas, you can plug in the numbers yourself:
d_{V} = Vsin \theta t + \frac{1}{2} at^2

0 = (Vsin \theta) t - \frac{1}{2} gt^2

0 = Vsin \theta - \frac{1}{2} gt

t = \frac{2Vsin \theta}{g}

this is the formula for total flight time, now you must use this formula to sove for distance:

d_{H} = Vcos {\theta}t

d_{H} = Vcos {\theta} (\frac{2Vsin \theta}{g})

d_{H} = \frac{V^2 sin2{\theta}}{g}

now just sub in your numbers and you should be ok.
 
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