A stone dropped from certain point crosses the top and bottom ends of a vertical pole of length 20m in one second the distance of top of the pole from the initial point of release will be (g=10)
Ans:11.25m
d = V*t + 0.5g*t^2.
20 = V*1 + 0.5*10*1^2,
V = 15 m/s = Velocity at top of pole.
V^2 = Vo^2 + 2g*d.
15^2 = 0 + 20d,
d = 11.25 m.
as it crosses pole
average speed = -20 m/s
change in speed in one second
= a(1) = -10 m/s
so
speed at top of pole = -15 m/s
so how far did it fall to get to -19.5
-15 = -10 t
t = 1.5
d= (1/2) g t^2 = 5(1.5)^2
19.361
h = (1/2)g t^2
= 5 * 2.25
= 11.25
S=vt + 1/2at²
20= v(1)+1/2(10)(1)²
20= v+5
v=15
v²=u²+2as
(15)²=(0)²+2(10)d
d=11.25
Why did the stone cross the top and bottom ends of the vertical pole? Because it wanted to reach new heights and explore new depths, of course! But let's do some calculations to figure out where it ended up.
Since the stone falls freely under the influence of gravity, we can use the equation:
distance = initial velocity × time + 0.5 × acceleration × time^2
In this case, the initial velocity is 0 (the stone is dropped), the acceleration is the acceleration due to gravity (10 m/s^2), and the time is 1 second.
So, the distance = 0 × 1 + 0.5 × 10 × 1^2 = 0 + 0.5 × 10 × 1 = 5 meters.
Now, we know that the length of the pole is 20 meters. Since the stone crossed both the top and bottom ends of the pole, we can divide the total distance traveled by the stone (which is 5 meters) by 2 to get the distance from the initial point of release to the top of the pole.
Therefore, the distance from the initial point of release to the top of the pole is 5 meters / 2 = 2.5 meters.
Hmmm, it looks like my calculations ended up in a different place than the provided answer. Maybe I dropped the ball (or, in this case, the stone). My apologies for any confusion caused!
To find the distance of the top of the pole from the initial point of release, we need to analyze the motion of the stone.
Let's break down the problem into two parts: the stone's motion from the initial point to the top of the pole and from the top of the pole to the bottom of the pole.
1. Stone's motion from the initial point to the top of the pole:
Since the stone is dropped, its initial velocity is zero. The only force acting on it is gravity. Using the equation of motion, h = ut + (1/2)gt^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time, we can find the height the stone reaches.
h = (1/2)gt^2
= (1/2) * 10 * (1^2) (As given, g = 10 and the time interval is 1 second)
= 5 meters
So, the stone reaches a height of 5 meters from the initial point to the top of the pole.
2. Stone's motion from the top of the pole to the bottom of the pole:
The stone is again falling freely due to gravity, starting from rest at the top of the pole and falling a distance equal to the height of the pole.
Using the equation h = ut + (1/2)gt^2, where the initial velocity is zero, we can find the time it takes for the stone to fall to the bottom of the pole.
h = (1/2)gt^2
20 = (1/2) * 10 * t^2
t^2 = 4
t = 2 seconds
Therefore, it takes 2 seconds for the stone to fall from the top to the bottom of the pole.
Now, to find the distance of the top of the pole from the initial point of release, we need to calculate the distance traveled by the stone in 1 second. We can use the equation d = ut + (1/2)gt^2, where u is the initial velocity (zero), g is the acceleration due to gravity, and t is the time.
d = 0 * 1 + (1/2) * 10 * (1^2)
= 0 + 5
= 5 meters
Therefore, the distance of the top of the pole from the initial point of release is 5 meters.
I apologize, but it seems there was an error in the given answer. The correct distance of the top of the pole from the initial point of release is 5 meters, not 11.25 meters.