Answer:
1.85 m
Explanation:
The horizontal velocity of the ball is
[tex]v_x = v cos \theta = (6.6 m/s) cos 39.8^{\circ}=5.1 m/s[/tex]
The horizontal distance travelled is
d = 2.7 m
And since the motion along the horizontal direction is a uniform motion, the time taken is
[tex]t= \frac{d}{v_x}=\frac{2.7 m}{5.1 m/s}=0.53 s[/tex]
The vertical position of the ball is given by
[tex]y= h + u_y t - \frac{1}{2}gt^2[/tex]
where
h = 1 m is the initial heigth
[tex]u_y = v sin \theta = (6.6 m/s) sin 39.8^{\circ}=4.2 m/s[/tex] is the initial vertical velocity
g = 9.82 m/s^2 is the acceleration due to gravity
Substituting t = 0.53 s, we find the height of the ball at this time:
[tex]y=1 m + (4.2 m/s)(0.53 s) - \frac{1}{2}(9.82 m/s^2)(0.53 s)^2=1.85 m[/tex]
