Solving Vector Diagrams: Finding Displacements & Velocities

[g9WfI]

Homework Statement

A stone is thrown horizontally at 15 m/s from the top of a vertical cliff, 50 m above the sea. Calculate the distance from the bottom of the cliff to the place where the stone hits the water. g = 9.8 m/s^2

Relevant Equations

trig, suvat

Here is my attempt at the vector diagram:

Could anyone give me any clues as to where to go from here? Is this diagram correct?
I tried finding θ using inverse tan 50/15 but I don't think I can do that because that's mixing up velocity and displacement.

EDIT: I copied and pasted the incorrect mark scheme here.
Correct mark scheme:
For the fall, s = 50 m, u = 0, v = ?, a = 9.8 m s–2 , t = ?
s = ut + ½at2 ⇒ 50 m = 0 + 4.9 m s-2 × t 2 ⇒ t 2 = 50 m ÷ 4.9 m s–2 = 10.2 s2
t = √(10.2 s2 ) = 3.19 s
horizontally, s = ut = 15 m s–1 × 3.19 s = 48 m (2 s.f.)

How do I find the corresponding displacements/velocities of the vectors with only this information?
I don't know if I'm making much sense, but I don't understand how to get the displacement from velocity;
I know s = vt but I'm not really getting anywhere with my working.

 

[Doc Al]

Not sure why you are attempting a vector diagram here. This is a projectile motion problem. One "secret" is to treat horizontal and vertical motions separately. First consider the vertical motion, since you have all the info you need to figure out the time it takes for the stone to hit the ground.

 

[bberisford]

This is a projectile motion problem. Split the problem into the horizontal (x) and vertical (y) directions. Analyze the projectiles motion in each direction independently using the kinematic equations.

 

[LeafNinja]

That diagram does not seem helpful.
You have already given the solution for the problem.

Velocity in the horizontal direction is constant.
Acceleration in the vertical direction is constant.
Time can be found by solving for time from the equation for vertical displacement, which you have given.
This can be used in the equation for horizontal distance to compute horizontal distance.

 

[haruspex]

You can't mix vectors of different physical dimensionality like that. It doesn’t mean anything to add a velocity and a displacement.

 

[Delta2]

Yes you are right that it is wrong to mix distance and velocity in $\tan\theta=\frac{50}{15}$, instead you should say that $\tan\theta=\frac{50}{48}$ as i think you calculated correctly the horizontal distance to be 48m.

If you want a vector diagram for velocity the main problem is that the vector of velocity keeps changing in magnitude and direction for this problem. However in cartesian coordinate system the vector of velocity formula is $$\vec{v}(t)=v_x\hat x+v_y\hat y=15\hat x+9.8t\hat y$$ as the x-component of velocity is constant and equal to 15 but the y-component of velocity keeps changing as time passes and it is $v_y=gt=9.8t$ where t the time elapsed.