Answer:
[tex]v_y=28.8m/s[/tex]
Explanation:
To simplify this problem we can just realize that the vertical velocity upwards on the top of the building will be equal to the vertical velocity downwards on the top of the building and start our calculations from here. The magnitude of this vertical velocity will be 15sin(53°) m/s.
We want then to calculate the final velocity of a stone with initial velocity 15sin(53°) m/s going downwards after traveling a distance d=35m downwards under the acceleration of gravity [tex]g=9.8 m/s^2[/tex], which is downwards. Since everything goes downwards that will be our positive direction.
Then we use, restricted on the vertical, the equation:
[tex]v_y^2=v_{0y}^2+2a_yd_y[/tex]
Then we have:
[tex]v_y=\sqrt{v_{0y}^2+2a_yd_y}=\sqrt{(15sin(53^{\circ}) m/s)^2+2(9.8m/s^2)(35m)}=28.8m/s[/tex]
The magnitude of the vertical velocity component of the rock as it hits the ground is 28.93 m/s.
The given parameters;
The time of motion of the stone is calculated as follows;
[tex]h = v_0_y t + \frac{1}{2} gt^2\\\\35 = (15 \times sin(53) )t + 0.5\times 9.8t^2\\\\35 = 11.98t + 4.9t^2\\\\4.9t^2 + 11.98t - 35 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\ a = 4.9, \ b = 11.98 , \ c = -35\\\\t = \frac{-b \ \ -/+ \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-11.98 \ \ -/+ \ \ \sqrt{(11.98)^2 - 4(4.9\times -35)} }{2\times 4.9}\\\\t = 1.72 \ s[/tex]
The magnitude of the vertical velocity component of the rock as it hits the ground is calculated as;
[tex]v_y_f = v_0_y + gt\\\\v_y_f = (15 \times sin(53)) + (1.73\times 9.8)\\\\v_y_f = 11.98 \ + \ 16.95\\\\v_y_f = 28.93 \ m/s[/tex]
Thus, the magnitude of the vertical velocity component of the rock as it hits the ground is 28.93 m/s.
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