Sitting at the edge of a clt, you throw a stone vertically pward with a velociy
ID: 2886359 • Letter: S
Question
Sitting at the edge of a clt, you throw a stone vertically pward with a velociy of 30 mis. The height of the stone above t the cifs edge in units of 10 meters can be approximated by 3t-0.5 (a) From the graph, at what time does the b) How high above the ciff does the stone reach? (Round your answer to one decimal place) c) When is the stone aganlat t nswer to one decimal place) d) You chack your watch and observe that the stone hits the ground b stone reach its highest point? (Round your answer to one decimal place.) the eilevation of the cliffs edge? (Round your a below the ciftf after 7 seconds How high is the ciff? (Round your answer to one decimal placeExplanation / Answer
(a) At heighest point, final velocity (v) =0, initial velocity (u) = 30 m/s (given)
v=u + at ( where a is accelaration due to gravity)
0 = 30 - 9.8 t ( putting a= 9.8 m/s^2)
t= 30/9.8 = 3.06 sec
(b) To reach at the highest point stone take 3.06 sec (from (a)).
Given that, f(t) = 3t -0.5 t^2 .................(*)
The highest distance above the cliff stone will reach after t = 3.06 sec,
Therefore, f(3.06)= 3 * 3.06 - 0.5 (3.06)^2= 4.49.
The highest point above the cliff = 44.9 m {4.49 *10= 44.9}
(c) From the highest point to the cliff edge, the time will take same.
Thus, the stone against the elevation of the cliffs edge will take 3.06 *2 sec= 6.12 sec.
(d) f(7)= 3*7 - 0.5 (7) ^2 ( putting t =7 in eq. (*))
f(7)= - 3.5
Thereore, the cliff is 35 m high from the ground.