$(20$ points$)$ A vertical curve has a curve length of $L=800ft$ and connects two grades $G_1=3\%$ and $G_2=-2\%.$ The point of vertical intersection $($PVI$)$ is at station $29+60\mathrm{.}00,$ with an elevation of $283\mathrm{.}2ft.$

a$.(4$ pts$)$ Find the station and elevation at the point of vertical curvature $($PVC$).$

Station $25+60\mathrm{.}00$

Elevation $271\mathrm{.}2ft$

b$.(4$ pts$)$ Find the station and elevation at the point of vertical tangency $($PVT$).$

Station $33+60\mathrm{.}0033+60\mathrm{.}00$

Elevation

$275\mathrm{.}2ft$

c$.(4$ pts$)$ Find the station and elevation of the highest point of the curve.

sation $30+40\mathrm{.}00$

Elevation

$278\mathrm{.}4ft$

d$.(4$ pts$)$ An overpass highway with an underside elevation of $291\mathrm{.}2ft$ will be built perpendicularly to the curve. The new highway must maintain at least a $16ft$ clearance over the existing curve. Find the range$($s$)$ of location $($shown in station$)$ on the vertical curve over which the new overpass can be constructed.

Any Location between PVC and $27+20\mathrm{.}00$ or beyond PVT

$(4$ pts$)$ A car is traveling downhill on this vertical curve, where grade equals $G_2=-2\%,$ at a speed of $35$ miles per hour. Determine the required stopping sight distance. $[$Use the perception$-$reaction time of $2\mathrm{.}5$ seconds and deceleration rate of $[$:$11\mathrm{.}2f\frac{t}{s^2}\}$

$ssD252\mathrm{.}9,$

Answers are given just need to know how to solve.