Question

. Chase is standing at the base of a 60-foot cliff. He throws a rock in the air hoping to get the rock to the top of the cliff.

If the rock leaves his hand 6 ft. above the base at a velocity of 80 ft/s, does the rock get high enough to reach the top
of the cliff? How do you know? If so, how long does it take the rock to land on top of the cliff (assuming it lands on the
cliff)? Graph the function and label the key features of the graph.

Answer 1

1. the rock gets high enough to reach the top of the cliff

2.We know from the solution given above in answer 1

3. 0.804secs to reach the top of the cliff

4. look up the graph in the attached file

Explanation:

Going by Newtons equation of motion , we have

h(t)=-16t^2+ut+h..............................1

juxtaposing into the equation we have

at maximum height

h(t)=the maximum height

t=time

u=initial velocity given as 80ft/s

a=acceleration due to gravity 32ft/s

h=initial height

since the rock is going against gravity, its going to be negative

dh/dt=o

differentiating 1 above we have

dh/dt=0=-32t+u

-80=-32t

t=2.5secs

to find the height it will get to at 2.5 secs will be

h(2.5)=-16(2.5)^2+80(2.5)+6

h(2.5)=106ft

1.the rock reaches a point higher than the cliff

2. we know from the solution given in 1

3.from the above equation

60=-16t^2+80(t)+6

54=-16t^2+80t

solving simultaneously

t=4.19s or 0.804secs

we go by 0.804secs