A projectile is shot from the edge of a cliff 145m above ground ...

[camila]

angle of 43 degrees with the horizontal.
a)Determine the time taken by the projectile to hit ground level
b)Determine the range X of the projectile as measured from the base of the cliff.
c)Find the maximum height above the ground reached by the projectile
d)Find the total velocity of the projectile at the maximum height above the ground.

[TI]

Damn, i learn this tommorow ='(

[Hawke]

list your varibles

vx = 60 * cos 43 m/s
vy = 60 * sin 43 m/s
s0 = 145 m

a)
first figure the time it takes to reach the apex.  at the apex vy = 0, therefore we know vf = vi + a*t and in vertical direction
0 = vy - g*t1

t1 = vy/g

figure the height of the projectile
from s = s0 + v0*t + 1/2at^2 in the vertical direction we get the following

s = s0 + vy*t - 1/2gt^2 (you know all the variables using t1 for t)

know that we know that s figure out how long it takes for it to reach the ground.  at the apex vy = 0 so we know
0 = s + 0 - 1/2gt2^2

t2 = sqrt (2*s/g)   (s was found a step earlier)

the total time it takes T = t1 + t2

b) range = T * vx

c) you already found that in part a)

d) at the max height the vertical component of the velocity is 0 so all you are left with is vh.