contestada

If a projectile is fired with an initial velocity ofv0 meters per second at an angleα above the horizontal and air resistance is assumedto be negligible, then its position after t seconds isgiven by the parametric equations below.
x = (v0cos(α))t
y = (v0sin(α))t - 4.9t2
(a) A gun is fired with α = 30° andv0 = 700 m/s. (Roundyour answers to the nearest whole number.)
(i) When will the bullet hit the ground?


(ii) How far from the gun will it hit the ground?


(iii) What is the maximum height reached by the bullet?


(b) Find the equation of the parabolic path by eliminating theparameter. (Use alpha for α and v_0 forv0.)

Respuesta :

Answer:

a)

(i) t  =  71 sec

(ii)   x (max)  = 43250 mt

(iii) y(max) = 6250 mt

b) y  =  x Tan α  -  4.9  x² /  V(o)² ( cos α) ²

Step-by-step explanation:

Equations describing a projectile motion are

x  =   V(o)*cos θ*t              y  = V(o)*sinθ*t  - gt²  

x(max)  =  V(o)²*sin2θ*t

a) α = 30°  and  V(o) = 700 m/sec

When  y =  0  means the projectile fall down to ground, the:

y = V(o) sin α * t  - 4.9 * t²      ⇒  y = 700 sin30 * t  - 4.9*t²

y = 700*(1/2)*t  - 4.9 t²      y  =  0     700*(1/2)*t  =  4.9 t²

700*(1/2)   =   4.9 t

t  = 71,43 sec        ⇒    t  =  71 sec

(ii) How far from the gun will it hit the ground .That is the point  of  x (max)

x (max)  =  V(o)²*sin2θ/g   x (max)  = (700)²(√3)/2*9,8

x (max)  =  (490000)*1.73/19.6     x (max)  = 43250 mt

(iii)  What is the mximum height

y = -1/2g*t²  + V(o)sin α*t + y(o)

y (max)  =  - 1/2 (9.8)*(71/2)²   (700)*1/2*(71/2)    ⇒ y(max) = 12425 - 6175

y(max) = 6250 mt

b)  We have  

x  =  V(o)  cos α * t      and   y =  V(o)  sin α * t  -  4.9* t²

From first equation    t   =  x / ( V(o)  cos α ) and by subtitution in second equation we have

y  = V(o)  sin α *  (  x / ( V(o)  cos α )  - 4.9   [ x / ( V(o)  cos α )]²

y  =  x Tan α  -  4.9  x² /  V(o)² ( cos α) ²