Respuesta :
Answer:
a) [tex]v_{o} =16m/s[/tex]
b) [tex]v=9.8m/s[/tex]
c) [tex]\beta =-35.46º[/tex]
Explanation:
From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial
Being said that, we can calculate the initial velocity of the ball
a) First we analyze its horizontal motion
[tex]x=v_{ox}t[/tex]
[tex]x=v_{o}cos(60)t[/tex]
[tex]v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}[/tex] (1)
That would be our first equation
Now, we need to analyze its vertical motion
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^2[/tex]
[tex]y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2[/tex]
Knowing [tex]v_{o}[/tex] in our first equation (1)
[tex]8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2[/tex]
[tex]\frac{1}{2}(9.8)t^2=16tan(60)-8[/tex]
Solving for t
[tex]t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s[/tex]
So, the ball takes to seconds to get to the other building. Now we can calculate its initial velocity
[tex]v_{o}=\frac{16m}{(2s)cos(60)}=16m/s[/tex]
b) To find the magnitude of the ball just before it strikes the building we need to calculate its x and y components
[tex]v_{x}=v_{ox}+at=16cos(60)=8m/s[/tex]
[tex]v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s[/tex]
So, the magnitude of the velocity is:
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s[/tex]
c) The direction of the ball is:
[tex]\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º[/tex]
The answers are as follows:
(a) the initial velocity is 16m/s
(b) the velocity just before the baseball strikes the building is 9.8m/s
(c) the direction o velocity just before the baseball strikes is -35.56° from the horizontal.
Given that the ball strikes the building 16m away and its final height is 8m more than the initial height.
Let v₀ be the initial velocity with an angle of θ = 60° from the horizontal.
So the horizontal velocity will be v₀cos60 and the verticle velocity will be v₀sin60
Now, the range:
r = v₀cos60×t
v₀ = r/tcos60
(a) Now the height h :
[tex]h=v_0sin60*t-\frac{1}{2}gt^2\\\\h=\frac{r}{t*cos60} sin60*t-0.5*9.8t^2\\\\8=16tan60-4.9t^2\\\\t\approx2s[/tex]
so the initial velocity:
[tex]v_0=r/tcos60\\\\v_0=16/2cos60\\\\v_0=16m/s[/tex]
(b) Now the horizontal velocity when the ball strikes:
[tex]v_x=v_0cos60\\\\v_x=8m/s[/tex]
and the verticle component:
[tex]v_y=v_0sin60-gt\\\\v_y=16sin60-9.8*2\\\\v_y=-5.7m/s[/tex]
so the resultant velocity:
[tex]v=\sqrt{v_x^2+v_y^2}\\ \\v=9.8m/s[/tex]is the velocity at which the ball strikes.
(c)The direction of the velocity with horizontal:
[tex]\alpha =tan^{-1}(v_y/v_x)\\\\\alpha =tan^{-1}(-5.7/8)\\\\\alpha=-35.46^o[/tex]
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