A projectile is fired with an initial speed of 75.2 m/s at an angle od 34.5 degrees above the horizontal on a long flat firing range. Determine (a) th

opatovaL

opatovaL

Answered question

2020-10-18

On a long flat firing range, a projectile is fired with an initial speed of 75.2 m/s at an angle of 34.5 degrees above the horizontal. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range) and (d) the velocity of the projectile 1.50 s after firing.

Answer & Explanation

Arnold Odonnell

Arnold Odonnell

Skilled2020-10-19Added 109 answers

Every time a projectile is fired with some velocity V atsome angle, you must break up the velocity into its x and ycomponents since velocity is a vector. The x component of thevelocity is Vx=Vcos(?), and the y component is Vy=Vsin(?). Now, in these range problems there are generally only twoequations that you will use. One equation will be for thex-component and the other will be for the y-component. Theseequations are X=Vx×T, (X=distance traveled in the x direection,T=total time), and Y=YVyT+(12)aT2.
Everytime you havethese equations two things happen. In the x-direction, the velocityVx is always constant since there are no forces acting in thisdirection. In the y-direction however, there is a force, gravity,which tends to act down towards the ground. Since we take anymotion moving up in the y-direction to be positive, gravity pointsdown and should be negative. So the equation in the y should be Y=YVyT(12)gT2
, where g is gravity. Now you are readyto solve the problem. We will answer b first. Since the projectileis fired on a long flat range, Yo=0 since there is no initial hillwith height Yo. Since at T=0 the object is on the ground gettingready to be fired and at T=? after being fired it lands again(Y=0), set 0=VyT(12)gT2 and solve for T, this is the total timein air. Now for part a, since the trajectory of the ballis symetrical, at exactly 1/2 the total time the ball will be atits maximum height. So cut the time you got for part b i half andplug it into the equation Y=VyT(12)gT2. For part c, thetotal distance covered deals with the X direction and so youshould take the total time of flight and plug that into theequation X=VxT. For part d, its a bit ridiculous and I willpost a picture of the solution of it after thispost.

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