Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 decade ago

Help in solving this physics question?

A stone is thrown vertically upwards with an initial speed of 10.0m/s from a cliff that is 50.0m high

1. When does the stone reach the bottom of the cliff?

2. What speed does it have just before hitting the ground ?

3. What is the total distance travelled by the stone ?

please could you give explanation for the answers if possible atleast tell the formulae with answer ... thanks a lot

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  • 1 decade ago
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    first I will address question 2. work done by gravity equals change in kinetic energy. (1/2)mv^2 + mgh=(1/2)mv^2 the LHS is the initial velocity and the RHS is the final velocity, notice that you do not know mass, but also notice that it cancels out, so

    (1/2)m(10^2)+m(9.81)(50)=(1/2)mv^2

    v=sqrt(100+(2)(50)(9.81))=32.88 m/s

    to answer the other 2 questions we break the problem into 2 parts. The stone going up and then coming down. first going up...

    we know that at its peak the velocity is 0, so this equation is usefull...

    V(final)=V(initial)+A*T......... 0=10-(9.81)T....T=1.02

    now that we know the time to the peak height we use this equation...

    D(final)=D(initial)+V*T+(1/2)A*T^2....

    D=0+(10)(1.02)+(1/2)(-9.81)(1.02^2)=5.1 m

    total distance traveled is then= (2)(5.1)+50 = 60.2 m

    now for the total time in the air....

    D(final) = -50

    -50=0+(10)*T+(.5)(-9.81)T^2

    solving that quadratic equation gives you T = 4.4 s roughly

    Source(s): University of Illinois (Kinetic Physics for Engineers)PHYS211
  • 1 decade ago

    just using kinetic equations for this problem.

    y(final)=y(initial)+v(initial)t+1/2at^2

    choose the origin on the ground

    y(final)=0

    y(initial)=50.0m

    v(initial)=10.0m/s

    a= -g = -9.81m/s^2 since the acceleration due to gravity points down and the coordinate we choose points up.

    then we have eq-n:

    0=50+10t-4.9t^2

    t=4.37 or t=-2.33 (dont take this negative solution)

    so after 4.37s, the stone will hit the ground or the bottom of the cliff

    2, using another kinetic eq-n for velocity

    v=v(initial)+at=10-9.81(4.37)= -32.87m/s (the negative sign indicate the stone is moving downward)

    3, i dont know if there is another way to calculate this part faster. Personally i'll calculate the highest point the stone will reach first

    use the equation v(final)^2-v(initial)^2 = 2*a*delta y

    at the top, the velocity of the stone is 0

    delta y = -10/(2*-9.81) = 0.51m

    so we have the total distance traveled by the stone = 0.51*2 + 50 = 51.02 m

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