Answer:
(a) X = 17.48 m
(b) Y = 8.903 m
(c) X = 28.33 m
(d) Y = 8.68 m
(e) X = 44.68 m
(f) Y = 0 m
Explanation:
Given;
magnitude of initial velocity, v = 21.1 m/s
angle of projection, θ = 39.9°
the horizontal component of the velocity, [tex]v_x = vcos \theta[/tex]
[tex]v_x = 21.1(cos 39.9^0) = 16.187 \ m/s[/tex]
the vertical component of the velocity, [tex]v_y = vsin \theta[/tex]
[tex]v_y = 21.1(sin 39.9^0) = 13.535 \ m/s[/tex]
(a) the horizontal components of its displacement;
[tex]X = v_xt + \frac{1}{2} gt^2[/tex]
gravitational influence on horizontal direction is negligible, g = 0
[tex]X = v_xt \\\\X = (16.187)(1.08)\\\\X = 17.48 \ m[/tex]
(b) the vertical components of its displacement;
[tex]Y = v_yt + \frac{1}{2} gt^2\\\\Y = (13.535*1.08) + \frac{1}{2} (-9.8)(1.08)^2\\\\Y = 14.618 -5.715\\\\Y = 8.903 \ m\\[/tex]
(c) the horizontal components of its displacement;
[tex]X = v_xt\\\\ X = (16.187)(1.75)\\\\X = 28.33 \ m[/tex]
(d) the vertical components of its displacement;
[tex]Y = v_yt + \frac{1}{2} gt^2\\\\Y = (13.535 *1.75) + \frac{1}{2} (-9.8)(1.75)^2\\\\Y = 8.68 \ m[/tex]
(e) the horizontal components of its displacement;
[tex]X = v_xt \\\\X = (16.187*5.09)\\\\X = 82.39 \ m\\[/tex]
(f) the vertical components of its displacement;
[tex]Y = v_yt + \frac{1}{2}gt^2\\\\ Y = (13.535*5.09)+ \frac{1}{2}(-9.8)(5.09)^2\\\\Y = -58.1 \ m[/tex]
this displacement is not possible because it is beyond zero vertical level; it shows that the stone will hit the ground before 5.09 s.
When the stone hits the ground Y = 0 m
At zero vertical displacement (Y = 0 m), the time at that position will be calculated as;
[tex]Y = v_yt+\frac{1}{2}gt^2\\\\0 = (13.535t) + \frac{1}{2}(-9.8)t^2\\\\0= 13.535t - 4.9t^2\\\\0 = t(13.535-4.9t)\\\\t = 0 \ \ or \\\\13.535-4.9t = 0\\\\4.9t = 13.535\\\\t = 2.76 \ s[/tex]
The horizontal displacement at this time is given by;
X = vₓt
X = (16.187 x 2.76)
X = 44.68 m
