a) The time of flight is 3.78 s
b) The initial speed is 17.0 m/s
c) The speed at impact is 46.4 m/s at [tex]70.3^{\circ}[/tex] below the horizontal
Explanation:
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a)
The motion of the ball is a projectile motion, which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction Â
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction Â
We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:
[tex]y=u_y t+\frac{1}{2}at^2[/tex]
where we have:
y = -45.0 m is the vertical displacement of the ball (the height of the building)
[tex]u_y=u sin \theta[/tex] is the initial vertical velocity, with u being the initial velocity (unknown) and [tex]\theta=-27.0^{\circ}[/tex] the angle of projection
t is the time of the fall
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity
Along the x-direction, the equation of motion is instead
[tex]x=(u cos \theta)t[/tex]
where [tex]ucos \theta[/tex] is the horizontal component of the velocity. Rewriting this equation as
[tex]t=\frac{x}{ucos \theta}[/tex]
And substituting into the previous equation, we get
[tex]y=xtan \theta + \frac{1}{2}gt^2[/tex]
And using the fact that the horizontal range is
x = 59.0 m
And solving for t, we find the time of flight:
[tex]t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s[/tex]
b)
We can now find the initial speed, u, by using the equation of motion along the x-direction
[tex]x=u cos \theta t[/tex]
where we know that:
x = 59.0 m is the horizontal range
[tex]\theta=-23^{\circ}[/tex] is the angle of projection
[tex]t=3.78 s[/tex] is the time of flight
Solving for u, we find the initial speed:
[tex]u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s[/tex]
c)
First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to
[tex]v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s[/tex]
The vertical velocity instead changes according to the equation
[tex]v_y = u sin \theta + gt[/tex]
Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:
[tex]v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s[/tex]
Where the negative sign means it is downward.
Therefore, the speed at impact is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s[/tex]
while the direction is given by
[tex]\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}[/tex]
So, [tex]70.3^{\circ}[/tex] below the horizontal.
Learn more about projectile motion:
brainly.com/question/8751410
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