2 Dimensional Motion and Vectors Help

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A stone is kicked off a 52 m cliff with an initial speed of 18 m/s. The time taken for the stone to reach the water is calculated to be 3.26 seconds using the equation for vertical motion. The initial calculation for the impact speed only considered the vertical component, yielding 31.98 m/s, which was incorrect. To find the correct impact speed, vector addition of the vertical and horizontal components must be performed. The correct speed at which the stone strikes the water is 36.889 m/s.
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Homework Statement



A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 18 m/s. The cliff is 52 m above the water's surface. How long does it take for the stone to fall to the water? With what speed does it strike the water?

Homework Equations


The Attempt at a Solution



So far, I was able to figure out the time:

dy= 1/2gt^2 where g=9.81 m/s^2
52m= 1/2*9.81*t^2
t= 3.26 seconds

I thought it would be correct to use Vy=gt to find the speed at which the stone strikes the water, therefore being (9.81 m/s^2)(3.26 s)= 31.98 m/s. But when i checked the answer key and talked to the teacher, she said that the answer was 36.889 m/s, and I was wrong.

So can someone tell me what I'm doing wrong and how to find the speed at which the stone strikes the water?
 
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You have only calculated the vertical component of the speed.

You need to do a vector addition of this with the horizontal component.
 
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