Answer:
54.0 m/s
Explanation:
We can start by calculating the time it takes for the ball to reach the ground. This can be done by using the equation for the vertical position of the ball at time t:
[tex]y(t) = h + u_y t - \frac{1}{2}gt^2[/tex]
where
h = 140 m is the initial height
[tex]u_y =0[/tex] is the initial vertical velocity (the ball is thrown horizontally)
g = 9.8 m/s^2 is the acceleration of gravity
When the ball reaches the ground, y=0, so we solve for t to find the time of flight:
[tex]0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(140)}{9.8}}=5.35 s[/tex]
We can find the vertical speed of the ball at the moment of the impact by using
[tex]v_y=u_y+gt[/tex]
Using t = 5.35 s,
[tex]v_y=0+(9.8)(5.35)=52.4 m/s[/tex]
The motion of the ball along the horizontal direction is a uniform motion - so the horizontal velocity is constant during the whole motion, and it is given by
[tex]v_x = \frac{d}{t}[/tex]
where
d = 69 m
is the distance travelled horizontally. Using t = 5.35 s,
[tex]v_x = \frac{69}{5.35}=12.9 m/s[/tex]
And therefore, the speed of the ball at the moment of the impact is given by
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(12.9)^2+(52.4)^2}=54.0 m/s[/tex]
