First, calculate the components of its initial velocity v0:
v0x = v0cos10 = 24.6 m/s
v0y = v0sin10 = 4.34 m/s
The ball reaches its peak when vy = 0. Let's calculate the time it takes for vy to become zero:
vy = v0y - gt ---> t = v0y/g = 0.44 s
The horizontal distance it travels in this time is
x = v0xt = (24.6 m/s)(0.44 s)
= 10.8 m
Note that the net is 15 m away. After traveling a horizontal distance of 10.8 m, the height of the ball is
y = -(1/2)gt^2 + v0yt + 2.1
= -(4.9 m/s^2)(0.44 s)^2 + (4.34 m/s)(0.44 s) + 2.1 m
= -0.95 m + 1.9 m + 2.1 m
= 3.05 m
Note that this is the height of the ball at its peak. While the ball is well above the net at its peak, it is well short of its required horizontal distance to clear it. Instead, let's find the time it takes for the tennis ball to travel a horizontal distance of 15 m first:
x = v0xt ----> t = x/v0x = (15 m)/(24.6 m/s) = 0.61 s
Then calculate the height y when t = 0.61 s. If y > 0.9 m (height of the net), then the ball will clear the net.
y = -4.9t^2 + v0yt + 2.1
= -4.9(0.61 s)^2 + (4.34 m/s)(0.61 s) + 2.1 m
= -1.82 m + 2.65 m + 2.1 m
= 2.93 m
Yes, the ball will clear the net by 2.03 m.
