Respuesta :
Emily throws the ball at 30 degree below the horizontal
so here the speed is 14 m/s and hence we will find its horizontal and vertical components
[tex]v_x = 14 cos30 = 12.12 m/s[/tex]
[tex]v_y = 14 sin30 = 7 m/s[/tex]
vertical distance between them
[tex]\delta y = 4 m[/tex]
now we will use kinematics in order to find the time taken by the ball to reach at Allison
[tex]\delat y = v_y *t + \frac{1}{2} at^2[/tex]
here acceleration is due to gravity
[tex]a = 9.8 m/s^2[/tex]
now we will have
[tex]4 = 7 * t + \frac{1}{2}*9.8 * t^2[/tex]
now solving above quadratic equation we have
[tex]t = 0.44 s[/tex]
now in order to find the horizontal distance where ball will fall is given as
[tex]d = v_x * t[/tex]
here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time
[tex]d = 12.12 * 0.44 = 5.33 m[/tex]
so the distance at which Allison is standing to catch the ball will be 5.33 m
The horizontal distance from the base of the dorm Allison should stand in order to catch the ball is 5.3 m.
The given parameters;
- angle of projection, θ = 30⁰
- horizontal speed of the ball, v = 14 m/s
- vertical distance, h = 4 m
The time of motion of the ball from the given height is calculated as follows;
[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\4 = (14 \times sin30)t \ + \ (0.5 \times 9.8)t^2\\\\4 = 7t + 4.9t^2\\\\4.9t^2 + 7t - 4 = 0\\\\a = 4.9, \ b = 7, \ c = -4\\\\t = \frac{b^2 \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{(7)^2 \ \ +/- \ \ \sqrt{(7)^2 \ -4(4.9\times -4)} }{2(4.9)} \\\\t = 0.437 \ s[/tex]
The horizontal distance traveled by the ball is calculated as follows;
[tex]X = v_0_xt\\\\X = 14 \times cos30 \times 0.437\\\\X = 5.3 \ m[/tex]
Thus, the horizontal distance from the base of the dorm Allison should stand in order to catch the ball is 5.3 m.
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