Kinematic Physics Problem?

Question

This is a problem i'm a bit confused with, partially cuz i dont know the golf term and not sure if the ball lands after the peak? Anyway, heres the problem..

A golfer faces an approach from the fairway to an elevated green 6 meters above where she stands. She uses a pitching wedge with a loft of 70degrees and the ball leaves her club traveling at 35 m/s.

a) how long is the ball in the air before it lands on the green?

b) how far is the green from where she is standing?

c) what is the maximum height the ball reaches above the level from which it was hit?

Answer 1

Angle with horizontal: =70 degrees

vector magnitude 35 m/s

Vx:

35* Cos(70)=11.97 m/s

Vy

35* Sin(70)=32.89 m/s

t=-32.89/-9.8 = 3.356 "seconds to max height"

c) max height =(1/2)*9.8*(3.356)^2 = 55.15 meters

fall=  55.15 - 6 = 49.15 meters " to green"

total flight time:sqr((49.15 * 2) / 9.8) + 3.356

a) = 6.52 seconds

b)  11.97*6.52 = 79.24  meters "horizontal"

vector distance=sqr(79.24^2+6^2) = 79.47 meters "slope"

Answer 2

I think what they are implying is that:

   - the ball's initial direction is 70 degrees from horizontal.

   - ball's initial velocity is 35 m/s

   - the ball lands 6 meters higher than it started.

   - the ball lands on the green

So, you break up that initial velocity into its components vx0 and vy0.

(draw the vector diagram, and figure vertical and horizontal part using sin's and or cosine).  

Then, solve the vertical equations like you would normally -- first find height & time where vertical velocity = 0 (using vy=v0y-g*t, solve for t, then plug into y=v0y*t - 1/2g*t^2).  This is the answer to part c.

Then find time to go from that height down to 6 meters (using y=y0 -1/2gt^2, where now y=6m, y0=height at top of arc, solve for t).  

Add these two times together, to get answer to a.

Then, figure out horizontal distance (x=v0x*t) using that total time to get answer to part b.