Answer:
110 m
Explanation:
First of all, let's find the initial horizontal and vertical velocity of the projectile:
[tex]v_{x0}=v cos 30^{\circ}=(25 m/s)(cos 30^{\circ})=21.7 m/s[/tex]
[tex]v_{y0}=v sin 30^{\circ}=(25 m/s)(sin 30^{\circ})=12.5 m/s[/tex]
Now in order to find the time it takes for the projectile to reach the ground, we use the equation for the vertical position:
[tex]y(t)=h+v_{0y}t-\frac{1}{2}gt^2[/tex]
where
h = 65 m is the initial height
t is the time
g = 9.8 m/s^2 is the acceleration due to gravity
The time t at which the projectile reaches the ground is the time t at which y(t)=0, so we have:
[tex]0=65+12.5 t - 4.9t^2[/tex]
which has 2 solutions:
t = -2.58 s
t = 5.13 s
We discard the 1st solution since its negative: so the projectile reaches the ground after t=5.13 s.
Now we know that the projectile travels horizontally with constant speed
[tex]v_x = 21.7 m/s[/tex]
So, the horizontal distance covered (x) is
[tex]x=v_x t = (21.7 m/s)(5.13 s)=111.3 m[/tex]
So the closest option is
110 m
