Projectile motion stone problem

[tigerlili]

Homework Statement

a stone is projected at a cliff of height h with an initial speed of 46.0 m/s directed at an angle θ0 = 65.0° above the horizontal. The stone strikes at A, 5.47 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.

Homework Equations

v0x= vcostheta
v0y= vsintheta

vfy= v0y-gt

The Attempt at a Solution

first, i broke the vector up into components
v0x= 19.44
v0y= 41.69

vfy= v0y-gt

y= -1/2gt^2 +v0yt + y0

then i tried to find b) the speed of the stone

i did vfy= v0y-gt
and got -11.92
that was wrong

i'm pretty lost when it comes to the other stuff.. please help?

 

[Fightfish]

The speed of the stone consists of both the vertical and horizontal components. After finding vfy, you have to perform a vector addition with vfx.
As with regards to the height of the cliff, just use a kinematics equation involving time, initial velocity, acceleration and displacement.
The maximum height of the stone occurs when the stone has instantaneous velocity of zero - can you see why this must be the case?

 

[tigerlili]

how is v0x calculated?

yes, i understand that the velocity at max height is 0 because it's the peak and the direction of the acceleration is changing

 

[Redbelly98]

how is v0x calculated?

We're given the initial velocity's magnitude and angle. From that you can use trig to get the x and y components, i.e. v_0x and v_0y.