Given data:
* The height of the stadium is 45 m.
* The horizontal velocity of the ball is 5.7 m/s.
Solution:
By the kinematics equation, the vertical motion of the ball is,
[tex]H=u_yt+\frac{1}{2}gt^2[/tex]where H is the height, g is the acceleration due to gravity, u_y is the vertical initial velocity, and t is the time taken,
The vertical initial velocity of the ball is zero.
Substituting the known values,
[tex]\begin{gathered} 45=0+\frac{1}{2}\times9.8\times t^2 \\ 45=4.9\times t^2 \\ t^2=\frac{45}{4.9} \\ t^2=9.2 \\ t=3.03\text{ s} \end{gathered}[/tex]Thus, the time taken by the ball to hit the ground is 3.03 seconds.
(b). By the kinematics equation, the horizontal motion of the ball is,
[tex]R=u_xt+\frac{1}{2}at^2[/tex]where u_x is the horizontal initial velocity, a is the acceleration, and R is the horizontal range,
The acceleration of the ball in the horizontal direction is zero.
Substituting the known values,
[tex]\begin{gathered} R=5.7\times3.03+0 \\ R=17.27\text{ m} \end{gathered}[/tex]Thus, the ball will land 17.27 meter from the base of stadium.
