Answer to Question #112488 in Mechanics | Relativity for Marc

Question

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

Answer 1

Solution.

$x = 86t;$ $y = 96t-4.9t^2;$

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

$x = \upsilon_0$_x$t;$ $y = \upsilon_0$_y $t + a_yt^2/2;$

a) $a_x = 0; a_y = -9.8m/s^2;$

b) $\upsilon_x = 86m/s ;$$\upsilon_y = 96-9.8t;$

c) $\upsilon_0 =\sqrt{\upsilon_x^2 + \upsilon_y^2 };$ $\upsilon_0 = \sqrt{86^2 + 96^2 } = 128.9 m/s;$

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

e) $h_m = -\upsilon_0$_y² /2$\sdot a_y;$

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

f) $L =υ_0 ​$_x$t_m;$ $t_m =2\sdot(-\upsilon$_0y$/a_y);$ $t_m = 2\sdot(-96/-9.8) =19.6s;$

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

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

b) $\upsilon_x = 86m/s ;$ $\upsilon_y = 96-9.8t;$

c) $\upsilon_0 = 128.9m/s;$

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

e)$h_m = 470.2m;$

f)$L = 1684.9m;$