Question #112488
A projectile fired at O follows a parabolic trajectory, given in parametric form by:

x = 86t and y = 96t – 4.9t2

where x and y are measured in meters and t in seconds. Determine:

(a) ax and ay at any time t

(b) vx and vy at any time t

(c) initial velocity, vo (velocity at pt O)

(d) velocity when t = 5s (hint: get first vx and vy at t = 5s)

(e) maximum height h

(f) horizontal distance L
1
Expert's answer
2020-04-27T10:00:08-0400

Solution.

x=86t;x = 86t; y=96t4.9t2;y = 96t-4.9t^2;

Since the trajectory of motion is a parabola, the motion of the projectile is described by the formulas:

x=υ0x = \upsilon_0xt;t; y=υ0y = \upsilon_0y t+ayt2/2;t + a_yt^2/2;

a) ax=0;ay=9.8m/s2;a_x = 0; a_y = -9.8m/s^2;

b) υx=86m/s;\upsilon_x = 86m/s ;υy=969.8t;\upsilon_y = 96-9.8t;


c) υ0=υx2+υy2;\upsilon_0 =\sqrt{\upsilon_x^2 + \upsilon_y^2 }; υ0=862+962=128.9m/s;\upsilon_0 = \sqrt{86^2 + 96^2 } = 128.9 m/s;

d) υx=86m/s;υy=969.85=47m/s;\upsilon_x = 86 m/s; \upsilon_y = 96 - 9.8\sdot 5 = 47m/s;

e) hm=υ0h_m = -\upsilon_0y2 /2ay;\sdot a_y;

hm=962/2(9.8)=470.2m;h_m = -96^2/2\sdot(-9.8) = 470.2m;

f) L=υ0L =υ_0 ​xtm;t_m; tm=2(υt_m =2\sdot(-\upsilon0y/ay);/a_y); tm=2(96/9.8)=19.6s;t_m = 2\sdot(-96/-9.8) =19.6s;

L=8619.6=1684.9m;L = 86\sdot19.6 = 1684.9m;

Answer: a) ax=0;ay=9.8m/s2;a_x = 0; a_y = -9.8 m/s^2;

b) υx=86m/s;\upsilon_x = 86m/s ; υy=969.8t;\upsilon_y = 96-9.8t;

c) υ0=128.9m/s;\upsilon_0 = 128.9m/s;

d) υx=86m/s;υy=47m/s;\upsilon_x = 86 m/s; \upsilon_y = 47m/s;

e)hm=470.2m;h_m = 470.2m;

f)L=1684.9m;L = 1684.9m;




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