This one is for two problems I am having a rough time with... Problem 1: A stunt

Question

This one is for two problems I am having a rough time with...

Problem 1: A stunt driver wants to make his car jump over eight cars parked side by side below a horizontal ramp.

Part A:

With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars, and the horizontal distance he must clear is 20 m.

The answer I got for this problem was 36 m/s

Part B: If the ramp is now tilted upward, so that "takeoff angle" is 14 above the horizontal, what is the new minimum speed?

I don't know this part.

Problem 2: A skier is accelerating down a 30.0 hill at 1.80 m/s2

Part A: What is the vertical component of her acceleration?

The answer I got was 0.900 m/s^2

Part B: How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is 212 m ?

I do not know how to find the answer for this part.

Explanation / Answer

PART A).

Horizontal distance travelled = V*t

V*t = 20
vertical distance travelled = u*t + 0.5*g*t^2
1.5 = 0 + 0.5*9.8*t^2
1.5 = 4.9*(20/V)^2
0.306 = 484/V^2
V = 36 m/s
PART B)------------------------------------------------------
horizontal velocity = v*cos14
vertical velocity = v*sin14

horizontal distance travelled = (v*cos14)*t
(v*cos14)*t = 20
vertical distance travelled = u*t + 0.5*g*t^2
1.5 = (-v*sin14)*t + 0.5*9.8*t^2
=> 1.5 = (-v*sin14)*[20/(v*cos14)] + 4.9* [20/(v*cos14)]^2
=> 1.5 =-4.98 + 2081/v^2
=> v =2.68 m/s

Problem 2)

from the triangle

sin theta =opposite/hypotunues

y/1.8 = sin 30

y = 1.8*sin30

y=0.9 m/s/s

vertical component of acceleration is 0.9 m/s/s downward

t = ?

V0 = 0 m/s

y = 212 m

ay = 0.9m/s/s

from kinematic equations

y = V0t + 1/2at2

212 = (0)(t) + 1/2(0.9)(t2)

(t2)=471

t=21.7 sec

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