5. A runaway truck (the driver has bailed) drives horizontally at 26.8 m/s (abou

Question

5. A runaway truck (the driver has bailed) drives horizontally at 26.8 m/s (about 60mph) off a vertical cliff that is h (meters) above
the sea. It is observed to hit the water 3.50 seconds later.
(i) Calculate h, the height of the cliff.
(ii) Calculate the vertical component of the velocity vector when the object hits the water.
(iii) Calculate the velocity of the object when it hits the water (Make sure you give the speed and direction, in other words, give the
velocity vector of the truck in polar form).

Explanation / Answer

use the kinimatic equations

y=(initial velocity in the y direction)*t+(1/2)gt^2

initial velocity in he y is 0 so we only consider teh (1/2)gt^2

y=(1/2)*9.8*3.5^2

y=60.025m <- height of the cliff

to find the velocity vector use another kinematics equation

(finial velocity)=(initial velocity)+g*t

initial velocity is 0

finial velocity= 9.8*3.5 = 34.3m/s

to find the direction use tan=(velocity in the y)/(velocity in the x)

tan=(34.3)/(26.8)

=51.998 degrees

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