a stone is dropped from the peak of a hill . IT covers the distance of 30m in last second of its motion. Find the height of the peak?
h-30=4.9(t-1)^2
h=4.9 t^2
set them equal, solve for t, time to fall
then solve for h
To find the height of the peak, we can use the equation of motion for an object in free fall.
Let's break down the information given:
Initial velocity (u) = 0 m/s (since the stone is dropped)
Distance covered in the last second (s) = 30 m
We need to find the height of the peak, which we can call "h."
The equation of motion we can use is:
s = ut + (1/2)gt^2
Where:
s = distance covered
u = initial velocity
t = time taken
g = acceleration due to gravity
Since the stone is dropped, the initial velocity is 0 m/s. Let's assume the total time it takes for the stone to reach the ground is "t" seconds.
Using the information given, we have:
30 = 0 * t + (1/2) * 9.8 * t^2 [using g = 9.8 m/s^2]
Multiplying through by 2 to eliminate the fraction:
60 = 9.8t^2
Rearranging the equation:
t^2 = 60 / 9.8
t^2 ≈ 6.13
Taking the square root of both sides:
t ≈ √6.13
t ≈ 2.48 seconds
Now, we can use this value of "t" to determine the height of the peak (h). We can use the equation:
h = (1/2) * g * t^2
Substituting the values:
h = (1/2) * 9.8 * (2.48)^2
h = 30.4 meters (approximately)
Therefore, the height of the peak is approximately 30.4 meters.