Answer:
122.5 m
Explanation:
To solve the problem, we need to know the speed at which the ball has been thrown: here I assumed it has thrown at a speed of
[tex]v=2.5 m/s[/tex]
The motion of the ball is a projectile motion, so it consists of two independent motions:
- A uniform motion along the horizontal direction
- A uniformly accelerated motion along the vertical direction
We know that the ball lands at a distance of
d = 12.5 m
away from the base of the building; so the time taken by the ball to cover this distance in the horizontal direction is
[tex]t=\frac{d}{v}=\frac{12.5}{2.5}=5 s[/tex]
The vertical distance covered by the ball is given by the suvat equation
[tex]s=\frac{1}{2}gt^2[/tex]
where
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Therefore, substituting t = 5 s,
[tex]s=\frac{1}{2}(9.8)(5)^2=122.5 m[/tex]
So, the height of the building is 122.5 m.
