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
- 1 decade agoFavorite Answer
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
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...
now that we know the time to the peak height we use this equation...
total distance traveled is then= (2)(5.1)+50 = 60.2 m
now for the total time in the air....
D(final) = -50
solving that quadratic equation gives you T = 4.4 s roughlySource(s): University of Illinois (Kinetic Physics for Engineers)PHYS211
- 1 decade ago
just using kinetic equations for this problem.
choose the origin on the ground
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:
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