As per the question Maya kicked the soccer ball at an angle of 37 off the ground.
The initial speed[u] given to the ball is 25 m/s
Resolving u into horizontal and vertical components we get
[tex]u_{x} =ucos37[/tex] and [tex]u_{y} =usin37[/tex]
[tex]u_{x}[/tex] is the horizontal component and [tex]u_{y}[/tex] is the vertical component.
The total time of ascent i .e the time taken by the ball to reach at the maximum height is given -
[tex]T_{as.}[/tex][tex]=\frac{usin\theta}{g}[/tex]
Where [tex]\theta[/tex] is the angle of projection with the ground and g is the acceleration due to gravity.
Here [tex]\theta[/tex]=37 degree
So [tex]T_{as.} =\frac{u\sin37}{g}[/tex]
=[tex]\frac{25*sin37}{9.8}[/tex]
=1.535242 second
As time of ascend is equal to time of descend, so total time of flight
T=3.070484 Second
As per the question she kicked the ball again at t=2.2 s later.
It is obvious from the above that the ball was downward in its trajectory.
At t = 2.2 s after she kicked the ball,the ball is still above and in a down condition
Parabolic motion is a combination of regular straight motion (x-axis) and straight motion changes irregularly (y-axis)
The formulas that apply include:
X-axis:
Vx = Vοx = Vο cos θ
X = Vοx t = Vο cos θ x t
Y-axis:
Vοy = Vο sin θ
Vy = Vοy - gt = Vο sin θ - gt
Y = Vy t - ½ gt²
While the time to reach the maximum / highest point is
Thigh = (Vo sin θ): g
While the time when reaching the maximum distance is 2 times when it reaches the highest point:
Tmax = 2. (Vo sin θ): g
It is known that Maya is kicking the ball with Vo = 25 m / s at an angle of 37° so that the time to reach its highest point:
T high = 25 sin 37/10
T high = 1.5045 seconds
And the maximum time when it reaches the ground again:
T max = 2. 1.5045 = 3.009 seconds
So when t = 2.2 s, the ball is still above and in a down condition
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the vertical velocity change
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Keywords : trajectory, ball, Parabolic motion,regular straight motion (x-axis),straight motion changes irregularly (y-axis)
