A rock is thrown straight down, not dropped, from the roof of a building that is 61 m above the ground. If it takes 3.1 s to rea
ch the ground, with what speed was it thrown?
Thank you
Thank you
1 answer:
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Answer:
v = √2G / R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1 / R = - G m1
/ R
v² = 2G (1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G / R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1 / R
Em = - G m1 / R
R = int ⇒ Em = 0
Answer:
option (a) 0.61 s
Explanation:
Given;
Time taken by the ball to reach the ground = 0.50 s
Let us first calculate the distance through which the ball falls on the ground
from the Newton's equation of motion, we have
where,
s is the distance
a is the acceleration
t is the time
here it is the case of free fall
thus, a = g = acceleration due to gravity
u = initial speed of the ball = 0
on substituting the values, we get
or
s = 1.225 m
Now,
when the elevator is moving up with speed of 1.0 m/s
the initial speed of the ball = -1.0 m/s (as the elevator is moving in upward direction)
thus , we have
or
or
4.9t^2 - t - 1.225 = 0
or
t = 0.612 s
hence, the correct answer is option (a) 0.61 s