Respuesta :
Answer:
[tex]V=\sqrt{V_{0}^{2}+2gy}[/tex]
Explanation:
Data given,
[tex]velocity,v =v_{0}\\ angle =\alpha _^{0}[/tex]
since the motion part is describe by a projectile motion, the acceleration along the horizontal axis is zero
Hence using the equation v=u+at we have the following equation ,
the velocity along the horizontal axis to be
[tex]V_{x}=V_{0}cos\alpha _{0} \\[/tex]
the velocity along the vertical axis to be
[tex]V_{y}=V_{0}sin\alpha _{0}-gt \\[/tex]
the magnitude of this velocity can be determine using Pythagoras theorem
[tex]V^{2}=V_{x}^{2} +V_{y} ^{2}[/tex]
if we substitute the expressions we have
[tex]V^{2}=V_{0}^{2}cos\alpha _{0}^{2} +(V_{0}sin\alpha _{0}-gt)\\expanding \\V^{2}=V_{0}^{2}(cos\alpha _{0}^{2}+sin\alpha _{0}^{2})-2gtsin\alpha _{0}+(gt)^{2}\\(cos\alpha _{0}^{2}+sin\alpha _{0}^{2})=1\\V^{2}=V_{0}^{2}-2gtsin\alpha _{0}+(gt)^{2}\\[/tex]
[tex]V^{2}=V_{0}^{2}-2gtV_{0}sin\alpha _{0}+(gt)^{2}\\V^{2}=V_{0}^{2}-2g(V_{0}sin\alpha _{0}-\frac{1}{2}gt^{2} )\\V_{0}sin\alpha _{0}-\frac{1}{2}gt^{2}=distance=y\\V^{2}=V_{0}^{2}-2gy\\ for upward \\V^{2}=V_{0}^{2}+2gy\\V=\sqrt{V_{0}^{2}+2gy}[/tex]
The speed of the rock, just before it strikes the ground (maximum speed) can be given as,
[tex]v=\sqrt{v_0^2-2gh} \\[/tex]
Witch is independent to angle [tex]a_0[/tex].
What is speed of the object just before hitting the ground?
The speed of the object falling from a height achieved the maximum speed, just before hitting the ground.
Given information-
The rock is thrown with a velocity [tex]v_0[/tex].
The rock is thrown with a angle of [tex]a_0[/tex].
The height of the building is [tex]h[/tex].
The height of the building can be given as,
[tex]h=v_0\times\sin(a_0)-\dfrac{1}{2}gt[/tex] .........1
Let the above equation as equation 1,
The horizontal velocity of the rock can be given as,
[tex]v_h=v_0\times\cos (a_0)[/tex].
The vertical velocity of the rock can be given as,
[tex]v_v=v_0\times\sin(a_0)-gt[/tex]
Here, [tex]g[/tex] is the gravitational force and [tex]t[/tex] is time.
Now the magnitude of the velocity can be given as,
[tex]v=\sqrt{v_h^2+v_v^2} \\v=\sqrt{(v_0\times\cos(a_0))^2+(v_0\times\sin(a_0)-gt)^2} \\v=\sqrt{v_0^2\times\cos^2(a_0)+v_0^2\times\sin(a_0)^2+(gt)^2-2gtv_0\times\sin(a_0)} \\v=\sqrt{v_0^2(\cos^2(a_0)+\times\sin(a_0)^2)-2g(v_0\times\sin(a_0)-\dfrac{1}{2}gt^2} \\v=\sqrt{v_0^2(1)-2g(v_0\times\sin(a_0)-\dfrac{1}{2}gt^2} \\[/tex]
Put the values of [tex]h[/tex] from equation 1 to the above equation as,
[tex]v=\sqrt{v_0^2-2gh} \\[/tex]
Hence the speed of the rock, just before it strikes the ground (maximum speed) can be given as,
[tex]v=\sqrt{v_0^2-2gh} \\[/tex]
Witch is independent to angle [tex]a_0[/tex].
Learn more about the velocity of free falling body here;
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