Trajectory question(find spped of impact)

[vande060]

A golfer drives a ball with initial angle
θ = 45 from a tee at the origin. The ball
hits a window at height h = 10m a distance
D = 80m away.

a. Find the initial launch speed v0 of the ball.

b. Suppose a second golfer tees off from the
same spot at an unknown angle and speed,
and miraculously her shot also hits the window
at the same height 1.8 s after launch. With
what speed did it hit the window?

c. Was her ball ascending or descending or in level flight when it hit?




y(x) = tan*x -.5*g*(x^2/v0^2*cos^2)



part a was easy enough, i solved the above trajectory equation for v0 to get v0=30m/s, but after that i can't figure out how to solve part b and c. it seems everything i could solve for to get the velocity of impact requires knowing the initial angle and initial speed

 

[vande060]

bump


i am revisiting this problem currently, still having trouble with parts b and c

 

[vande060]

all right, if this method does not have any critical errors in it, i believe i solved it(fingers crossed)so this is how i did it:

since t =1.8seconds, i can find out the product of cosθ and v0

x(t) = v0*cosθ*t

80/1.8 = v0*cosθ

44.45= v0*cosθ

i can do similar for finding the product of v0 and sinθ from the formula

y(t) = v0*sinθ*t - 1/2*g*t^2

10 = v0*sinθ(1.8) - (1/2*9.8*1.8^2)

10 + 15.88 = v0*sinθ(1.8)

14.38 = v0*sinθ

then i proceeded to use these figures in the velocity functions..

vx(t) = v0*cosθ and vy(t) = v0*sinθ - gt

first the horizontal velocity is just the same:44.45= vx(t)

next, solving for vertical velocity

vy(t) = v0*sinθ - gt

= 14.38 - (9.8*1.8)

vy(t)= -3.26 which is the answer for c, because it is negative acceleration, meaning heading back to Earth i think

next the speed of impact is given by formula:
s = |v| = (vx^2+ vy^2)^1/2

so [(-3.26)^2 + (44.45)^2 ] = 44.57 m/s, which is the answer to b

is it right? i hope so, but i am unsure of some of my math here