A projectile is launched from ground level to the top of a cliff ...

[aero]

A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high. If the projectile lands on top of the cliff 6.8s after it is fired, find the initial velocity of the projectile ( (a)magnitude and (b)direction ). Neglect air resistance.

[curlz]

2D Kinematics so split into X and Y components:

X direction:
x = 195m
Vix = Vicos(theta)
Ax = 0 m/s^2
t = 6.8s

Y direction:
y = 135m
Viy = Visin(theta)
Ay = -9.8 m/s^2


Then we make two equations... one in x direction and one in y direction
You can isolate for a single variable.
(1) x = Vi*cos(theta)*t
    x/tcostheta = Vi

(2) y = Vi*sin(theta)*t + 0.5 a*t^2

substitute (1) into (2) and we get:

(3) y = [x*sin(theta)*t]/[t*cos(theta)] + 0.5 a*t^2
you see that the time cancels and sin/cos is tan so we can simplify to

(3) y = x*tan(theta) + (1/2)a*t^2
plug in your numbers and solve for theta

theta = 61.66 degrees which is your direction then we could plug that back into (1) to solve for Vo.

(1) Vo = x/t*cos(theta) = (195) / (6.8)cos(61.66)
Vo = 60.4 m/s

Hope it's right!

[Jesse_J]

Height of Cliff = 135   = Vo Sin theta (t) - 1/2 g t^2   = Vo Sin theta (6,8)  - 4.9 (6.^2
                                                                             135   = 6.8 Vo Sin theta - 227
                                                                              362   =6.8 Vo Sin theta
                                                                               362/6.8 =  Vo Sin theta   = 53.2
Horizontal motion                                                                      Vo Cos theta = 195
                                                                                                    Tan theta   = 53.2/195  and theta = 17 degrees
VoCos theta (6,8)   = 195      
Vo  Cos theta    =   28.7 m/s
Vo   =    29.75 n/s
Vo Sin theta = 7.85 m/s