To find the angle she hits the water at we need her final velocity vector, as the angle that this makes with the horizontal is the angle that she will hit the water at.
At the point where she leaves the slide, her vertical velocity will be dependent on her acceleration due to gravity, while her horizontal velocity will be constant. As you know that the angle of the slide is
, then the direction of her velocity vector at the point where she leaves the slide is in a direction such that it make a
angle with the horizontal. So you can use this
as the initial velocity, splitting it into horizontal and vertical components.
 & =13.68\cos\left(45^{\circ}\right)\underset{\sim}{i}-13.68\sin\left(45^{\circ}\right)\underset{\sim}{j}\end{alignedat})
Now, we can just look at the vertical acceleration and vertical velocity as these components will follow linear motion. Taking up as positive.
Her initial vertical velocity,
)
m/s
Falling under the influence of gravity,
m/s^2
The time taken for her to hit the water from the end of the slide,
s
Now we need her final vertical velocity.
+-9.8\left(0.27\right)<br />\\ & =-12.319\text{ m/s}<br />\end{alignedat})
We now have the two velocity components at the point where she hits the water (remembering that the horizontal component will be the same as before because there is no force acting on her horizontally, only vertically).
That is her velocity when she hits the water is
 & =9.673\underset{\sim}{i}-12.319\underset{\sim}{j}\end{alignedat})
The angle that this vector makes with the horizontal is given by the tangent of the vertical component on the horizontal component. Since we are basically dealing with a right angle triangle with the opposite side to the angle being the vertical component and the adjacent side to the angle being the horizontal component, we can just look at the magnitude of the components.
 & =\frac{12.319}{9.673}<br />\\ \theta & =\tan^{-1}\left(1.2735\right)<br />\\ & =51.9^{\circ}<br />\end{alignedat}<br /> )
Home
Help
Search
Login
