Anonymous

# 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

Relevance

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
• 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