Answer: 11.025 meters.
Explanation:
Ok, we know that the initial velocity is only horizontal, so it does not affect the vertical problem.
Looking only at the vertical problem (because we want to know how high is the bridge) we have that:
The acceleration is the gravitational acceleration, g = 9.8m/s^2
A(t) = -9.8m/s^2
Where the negative sign is because this acceleration is downwards.
For the vertical velocity we integrate over time, as we do not have an initial vertical velocity, there is no constant of integration.
V(t) = (-9.8m/s^2)*t
For the position we integrate over time again, here the constant of integration is the initial vertical position, H, that is the height of the bridge.
P(t) = 0.5* (-9.8m/s^2)*t^2 + H.
Now we know that the ball hits the ground 1.5s after it was kicked, then:
p(1.5s) = 0m
With that we can find the value of H.
0 = 0.5* (-9.8m/s^2)*(1.5s)^2 + H.
H = 0.5*(9.8m/s^2)*(1.5s)^2 = 11.025m
