Two motion questions?

Question

1) An arrow is fired vertically upward at 40m/s from the top of a sheer cliff. It takes 12s to reach the base of the cliff. How high is the cliff?

2) How do I get distance from a velocity-time graph?

Thanks!

Answer 1

1)time af ascending = time of descending12/2= 6sec
then h =Vot-1/2gt^2 = 40x6 - 1/2 x9.8 x36=63.6 m

2) as we know that v =ds/dt ,then ds = vdt
then ∫ds = ∫vdt =area from t1 to t2
then s = area under v-t graph

Answer 2

h = 1/2 gt^2, where h is the height of the cliff and t is the time of fall starting at the cliff's edge. T = 12 sec = t + (time up + time down); where time up and down are the times the arrow took to go to apex and return to bow level. Time up = time down; so 2*time up is that extra time and T = t + 2*time up.

Time up can be found from v = u + g*time up; so that time up = u/g; where u = 40 m/sec and g = 9.81 m/sec^2 ~ 10 m/sec^2. Then time to fall from cliff's edge t = T - 2*time up = 12 - 2u/g ~ 12 - 80/10 = 4 sec. That is, of the 12 sec elapsed time between firing the bow and impact at the base of the cliff, 8 sec were spent going up then down from the bow and back. Only 4 seconds were spent in falling from the cliff's edge to its base.

Then from h = 1/2 gt^2 = 1/2 10*(4)^2 = 5 * 16 = 80 meters or there abouts. [NB: I used g ~ 10 m/sec^2 to simplify the math, use 9.81 m/sec^2 for more precise work. Also note I assumed the bow was at cliff height when fired. In a real case, the bow would be above cliff height in that the height of the person, plus arm extension should be considered.]

Answer 3

I. arrow....

hints:

break the motion into 3 parts.
a) up to the apex of the flight
b) then down to the original starting point at the cliff
c) then from the cliff to the ground

the time for (a) = time (b)

velocity at start of (a) = velocity when arrow reaches cliff in (b)

velocity at apex = 0

solution:

(a)

Vf = Vo + at
Vo = 40 m/s
Vf = 0
a = -9.8 m/s^2

so t = (40 m/s ) / (9.8 m/s^2) = 4.1 sec

(b) t from apex to cliff = t from cliff to apex = 4.1 sec

(c) since 12 s total and arrow has traveled 8.2 seconds so far, the arrow travels an addition 3.8 seconds from cliff to ground. also note that the arrow is traveling 40 m/s down at the start of this section.....

so you know Vo, t and a, calculate height of cliff

h = Vot + 1/2 at^2 = (40 m/s)x(3.8 s) + (1/2) x (9.8 m/s^2) x (3.8 s)^2 = 223 meters

II. distance from velocity / time graph

the first answerer was correct

v = ds/dt
vdt = ds
∫vdt = ∫ds

s = ∫vdt = area under curve of velocity on one axis and time on second axis

ie. integrate the curve in your graph.


**** update ****

the first answer was incorrect to assume time up = time down . he probably thought the arrow was fired up from the base of the cliff

the second answerer forgot to consider the velocity of the arrow at the cliff whilst on its way down. he assumed the velocity was 0 it was in fact 40 m/s