Respuesta :
Answer:
The initial speed of the soccer ball is 16.38 m/s
Explanation:
given;
vertical distance y = 2.44 m
horizontal distance x = 10.0 m
angle of projection θ = 25.0°
Initial velocity has two components, Vₓ and V[tex]_y[/tex]
Vₓ = V[tex]_i[/tex] cosθ
V[tex]_y[/tex] = V[tex]_i[/tex] sinθ
The horizontal distance = x = Vₓt + ¹/₂ ˣ g ˣ t², but g =0
x = Vₓt = V[tex]_i[/tex] cosθ *t
10 = V[tex]_i[/tex] cos25 *t
10 = 0.906V[tex]_i[/tex]*t
V[tex]_i[/tex]*t = 10/0.906 = 11.038 m
The vertical distance (g = - g, because it upward motion against gravity)
y = V[tex]_y[/tex]*t -¹/₂ ˣ g ˣ t²
2.44 = (V[tex]_i[/tex] sinθ)t - ¹/₂ ˣ 9.8 ˣ t²
2.44 = (V[tex]_i[/tex]*t)sinθ - ¹/₂ ˣ 9.8 ˣ t²
2.44 = (11.038)sin25° - 4.9t²
2.44 = (11.038)*0.4226 - 4.9t²
2.44 = 4.6647 - 4.9t²
4.9t² = 4.6647 - 2.44 = 2.2247
t² = 2.2247/4.9
t² = 0.454
t = √0.454
t = 0.674 s
Recall that V[tex]_i[/tex]*t = 11.038 m
V[tex]_i[/tex]*0.674 = 11.038 m, solve for V[tex]_i[/tex]
V[tex]_i[/tex] = 11.038/0.674
V[tex]_i[/tex] = 16.38 m/s
Therefore, the initial speed of the soccer ball is 16.38 m/s
The initial speed of the soccer ball is 16.38 m/s
given information:
vertical distance y = 2.44 m
horizontal distance x = 10.0 m
angle of projection θ = 25.0°
horizontal equation of motion:
The horizontal distance is given by:
since there is no force in the horizontal direction.
[tex]x = V_xt = Vcos\theta \times t\\\\10 = Vcos25 \times t\\\\10 = 0.906Vt\\\\Vt = 11.038 m\\\\[/tex]
vertical equation of motion:
The vertical distance is given by:
[tex]y = V_yt -\frac{1}{2} gt^2\\\\2.44 = (Vsin\theta)t - 0.5\times9.8t^2\\\\2.44 = (11.038)sin25 - 4.9t^2\\\\2.44 = (11.038)\times0.4226 - 4.9t^2\\\\2.44 = 4.6647 - 4.9t^2\\\\4.9t^2 = 4.6647 - 2.44\\\\t^2 = 2.2247/4.9\\\\t^2 = 0.454\\\\t = 0.674 s\\\\[/tex]
Now,
[tex]Vt = 11.038\\\\V\times0.674 = 11.038\\\\V = 11.038/0.674\\\\V = 16.38 m/s[/tex]is the initial speed of the soccer ball.
Learn more about equations of motion:
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