Question

A stone thrown off a bridge 20 m above a river has an initial velocity 14m/s of at an angle of 45 degrees above the horizontal. What is the range of the stone? At what velocity does the stone strike the water?

Answer #1

a)

You need to solve for time first using

yf = yi + visin?t + 1/2gt^2

0 = 20m + 14sin45t + 1/2(-9.8)t^2 and use the quadratic equation
to solve for t

t = 3.269 sec

To solve for the distance traveled use

xf = xi + vicos?t + 1/2at^2 there is no acceleration in the x
direction so that cancels

xf = 14cos(45)*3.269= 32.36m

b) For b I'm not sure if you what direction you want the final velocity in the x, y, or the direction its traveling so I'll just give all 3.

Theres no change in the velocity in the x direction so its just vfx = vixcos? = 14cos45 = 9.90m/s

For the y direction its vfy^2 = viy^2 + 2g(?y)

vfy = sqrt((14sin(45))^2 + 2(-9.8)(0-20m)) = 22.14m/s

The velocity the direction the stone is traveling is vf = sqrt(vx^2 + vy^2) = sqrt(9.90^2 + 22.14^2) = 24.25m/s

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