A car runs directly off a cliff with an initial velocity of 3.5 m/s. a) What are

Question

A car runs directly off a cliff with an initial velocity of 3.5 m/s.

a) What are the components of this velocity?

b) What will be the horizontal velocity 2 seconds after the car leaves the cliff?

c) If the cliff is 300 m high, at what time will the car reach the ground?

d) How far from the cliff will this car land?

e) If there is a small pond which begins 25m away from the cliff and extends 2.5 meters from there; will the car land in the pond?

f) What is the final vertical velocity at which the car is traveling? [The vertical velocity at the time when the car reaches the ground]

g) What is the final horizontal velocity at which the car is traveling? [The horizontal velocity at the time when the car reaches the ground]

h) What is the total final velocity of this motion? [magnitude and direction]

Explanation / Answer

part a:

initial horizontal speed=3.5 m/s

initial vertical speed=0 m/s

part b:

horizontal acceleration=0

so horizontal velocity will remain same through out

so after 2 seconds, horizontal velocity is 3.5 m/s

part c:

let time taken be t seconds.

vertical distance to be covered=300 m

vertical acceleration=9.8 m/s^2

using the formula:

distance=initial velocity*time+0.5*acceleration*time^2

==>300=0*t+0.5*9.8*t^2

==>t=7.8246 seconds

part d:

distance travelled horizontally=horizontal speed*time taken

=3.5*7.8246=27.3861 m

hence the car will land at a distance of 27.3861 m from the cliff.

part e:

yes, the car will land in the pond as pond stretches from 25 to 27.5 m

and the car lands at 27.3861 m

part f:

vertical velocity=initial vertical velocity+vertical acceleration*time

=0+9.8*7.8246=76.6811 m/s

part g:

final horizontal speed=3.5 m/s

part h:

final velocity =sqrt(horizontal speed^2+vertical speed^2)

=76.761 m/s

angle below horizontal=arctan(76.6811/3.5)=87.3866 degrees

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