Answer:
Yes, ball will clear the net
Explanation:
First we have to find the range of projectile motion.
Data given,
Ф = 7°
Initial velocity = 18 m/s
R = (V)^2.sin2Ф/g
Now by putting values
R = 7.99 m
Now for height
h = v^2.(sinФ)^2/2g
by putting values
h = 0.245 m
Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.
So, now from the maximum height, we have to calculate the horizontal distance of ball to net.
Now velocity in projectile motion is in two dimensions.
V(x) = 18 m/s
V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)
Now, by formula S = V(y)t + (1/2)gt^2
we can calculate time which is required by the ball to reach net from the maximum height it has achieved.
Now, tricky part is to calculate S, because without it we can not calculate t.
So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.
So, the vertical distance downward, which ball will travel from maximum height to net will be
S = 1.8 + 0.245 - 1.07
S = 0.975 m
Since we know V(y) = 0 m/s
S = (1/2)gt^2
t = (2S/g)^(1/2)
t = 0.44 s
Now time for both vertical and horizontal distance are same,
So, for horizontal distance "D(x)"
D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)
D(x) = 18 x 0.44
D(x) = 8.029 m
Now please notice that at maximum height, range was half, so at that point ball covered distance "a"
a = 3.99 m
From maximum height to net, as we calculated, ball covered
D(x) = 8.029 m
So, total distance covered by ball
a + D(x) = 3.99 + 8.029
a + D(x) = 12.024 m
which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.
