Refer to the figure shown below.
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m