a stone is dropped from a cliff which reaches the ground in8sec calculate the final velocity of stone g=9.8m/s
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Answers
Answer:
ans ... 78.4. m/s
Explanation:
use first equation of motion
u is 0 so
v=0+9.8*8
v =78.4
Answer:
The acceleration due to gravity is ~ 9.8 m/ss, let’s just use 10 for simplicity.
We can use the kinematic equations to find the answers to your question.
However, we are assuming here that the rock is falling in a vacuum.
Firstly the velocity 3 seconds later we can use:
V(f) = V(I) + at
f for final, i for initial, a acceleration, t time
V(f) = 0 + 10 * 3
= 30 m/s
Then for the second part we can use this equation to find the distance traveled in the 4th second only:
S = V(i) * t + (1/2) * a * t(^2)
Where s is the distance traveled
S = 30 * 1 + 1/2 * 10 * 1
= 35 meters!
Remember this is the disctace in the 4th second, not the total distance over all time. (80 meters is the total if you were wondering)
Of course this does not take into consideration air resistance, the shape and mass of rock. Lighter less aredynamic rocks would travel slower. Also we can consider the terminal velocity of an object. Where the upwards force of the air resistance is equal to the gravitational force on the mass.
A quick google gives a small rock like object a terminal velocity of aroumd 100–200mph, or 50–100 m/s, so only after 5 to 10 seconds of falling would a terminal velocity be reached.
It is very important to note that in a vacuum the only factor contributing to a change in velocity of a falling object is the gravitational pull. Shape or mass is irrelevant!
Hope that’s clear!