A caver slides off a rock at an angle of 28 deg below the horizontal and 21 m ab

Question

A caver slides off a rock at an angle of 28 deg below the horizontal and 21 m above the floor of the cave. Initial velocity of caver is 20 m/s. The ground below the rock is covered with sharp rocks, but there is a river 9.1 m away from the foot of the rock. a) Will he land in the river or on sharp rocks? b)What is the magnitude of his velocity when he impacts? c)At what angle (with respect to the horizontal) does he impact? Please show steps. I started a) by y=y(o)+V(yo)+1/2at^2 and then did quadratic to solve for time (I have 0.866 sec). Not sure if Im on right track?? Thank you in advance!

Explanation / Answer

A)

Initial velocity in the x direction:

20*cos(28) = 17.66 m/s

Initial velocity in the y direction:

-20*sin(28) = -9.39 m/s

Time it takes to fall:

d = Vi*t + (1/2)a*t^2

-21 = -(9.39)t - (1/2)(9.8)t^2

Solve for t:

t= 1.323 seconds

Distance covered in that time (horizontally):

d = Vi*t

d = 17.66*1.323 = 23.36 m

So, he will land in the river.

B)

Find final velocity in the y-direction:

Vf = Vi + at

Vf = -9.39 - (9.8)(1.323) = -22.36 m/s

Now combine the components of velocity to find magnitude:

|V| = sqrt((-22.36)^2 + (17.66)^2) = 28.49 m/s

C)

We know that the angle will be between 0 and 90 degrees below the horizontal.

Let's do it like this:

Arccos(17.66 / 28.49) = 51.69 degrees

So it is 51.69 degrees below the horizontal.

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