Kinematics: Solving Questions on Speed, Distance, Time & Acceleration

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A stone thrown vertically upward from a 79.0 m cliff with an initial speed of 11.0 m/s takes approximately 5.29 seconds to reach the bottom, with a speed of 40.9 m/s just before impact, covering a total distance of 91.4 m. A car rolling off a cliff from rest takes time to reach 74 km/h, while a sports car accelerates from rest to 99 km/h in 6.2 seconds, requiring calculation of average acceleration. Participants in the discussion are attempting to solve these kinematics problems but are struggling with the first question, leading to varied responses on the correct answers. Clarifications are sought regarding the calculations and the interpretation of speed and distance in the context of the problems. The discussion emphasizes the importance of setting up the correct equations of motion for accurate solutions.
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1.A stone is thrown vertically upward with a speed of 11.0 m/s from the edge of a cliff 79.0 m high (Fig. 2-32).

(a) How much later does it reach the bottom of the cliff?

(b) What is its speed just before hitting?

(c) What total distance did it travel?

2.If a car rolls gently (v0 = 0) off a vertical cliff, how long does it take it to reach 74 km/h?

3.A sports car accelerates from rest to 99 km/h in 6.2 s. What is its average acceleration in m/s2?
 
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Most of these can be solved once you set up an equation of motion representing your situation.

I suggest you make an attempt :o
 
yes i figured out the last two but the first one i am having trouble on... I found most of the missing variable but i can't seem to get the ones i need right.
 
the answers i have gotten for the first one is
A) 5.12
B)-39.3
C)-91.3
Can someone tell me if i did this wrong and help me
 
jimlee4262 said:
the answers i have gotten for the first one is
A) 5.12
B)-39.3
C)-91.3
Can someone tell me if i did this wrong and help me

a) 0.13424051
b)588.495976
s)79.0
 
the pro said:
a) 0.13424051
b)588.495976
s)79.0

a) 0.13424051s doesn't leave much time to reach the maximum height and then fall 79m plus the additional height achieved by the 11.0m/s initial velocity.
b) Accelerating to 588.495976m/s in 0.13424051s would require that acceleration due to the force of gravity is 4380.1979 m/s2
c) What about the distance upward due to the initial velocity of 11.0m/s?
 
I came up with
a) 5.29 s
b) 40.9 m/s (the problem asks for speed, which is the absolute value of velocity)
c) 91.4 m
 
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