Answer :
Answer:
[tex]y_{o}=213m[/tex]
Explanation:
From the exercise we know that the child hits the ball with an initial velocity, its direction and where it hits the ground.
First of all, we need to calculate how long does it take to hit the ground
[tex]x=v_{ox}t[/tex]
knowing that x=63m
[tex]63m=(13cos(51)m/s)t[/tex]
[tex]t=\frac{63m}{13cos(51)m/s}=7.70s[/tex]
Now, from free falling object's formula we know that position is:
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^2[/tex]
When the ball hits the ground it is at y=0m
[tex]0=y_{o}+(13sin(51)m/s)(7.70s)-\frac{1}{2}(9.8m/s)(7.70s)[/tex]
[tex]y_{o}=(4.9m/s^2)(7.70s)^2-(13sin(51)m/s)(7.70s)=213m[/tex]
So, the building is 213m tall
Height of the building
The height would be as follows:
[tex]213 m[/tex]
Given that,
The angle of kicking soccer ball [tex]= 51[/tex]°
Initial Velocity [tex]= 13 m/s[/tex]
Distance of the wall [tex]= 63m[/tex]
To find the time taken by the ball to reach the ground,
[tex]x = v_{ox}t[/tex]
so,
[tex]63m = 13 cos(51) m/s[/tex][tex])t[/tex]
∵ [tex]t = 7.7 sec.[/tex]
Now using the formula for free fall items,
[tex]y = y_{o} + v_{oy}t + 1/2 gt^2[/tex]
⇒ [tex]y_{o} = 4.9(7.7) - (13 sin(51)(7.7)[/tex]
∵ [tex]y_{o} = 213m[/tex]
Thus, [tex]213m[/tex] is the correct answer.
Learn more about "Initial Velocity" here:
brainly.com/question/9163788
