a body is dropped from rest at a heigth of 80m above the ground. (1)what is the speed just before it hits the ground?.(2)how long does it take to reach the ground?.
Just give me d answer
To find the answers, we can use the laws of motion and the equations of motion. Let's break down the problem into two parts:
(1) To find the speed just before the body hits the ground, we can use the equation of motion:
v = √(u^2 + 2as)
Where:
- v is the final velocity
- u is the initial velocity (which is 0 in this case since the body is dropped from rest)
- a is the acceleration due to gravity (approximately -9.8 m/s^2, taking negative due to the direction)
- s is the distance traveled (which is 80 m in this case)
Plugging in the values we have:
v = √(0^2 + 2 * (-9.8) * 80)
v = √(0 + (-1568))
v = √(-1568)
Since velocity cannot be negative in this context, we take the magnitude of the value:
v = √1568
v ≈ 39.6 m/s
Therefore, the speed just before it hits the ground is approximately 39.6 m/s.
(2) To find the time it takes to reach the ground, we can use another equation of motion:
s = ut + (1/2)at^2
Where:
- s is the distance traveled (which is 80 m in this case)
- u is the initial velocity (which is 0 in this case)
- a is the acceleration due to gravity (-9.8 m/s^2, taking negative due to the direction)
- t is the time
Plugging in the values we have:
80 = 0 * t + (1/2) * (-9.8) * t^2
80 = (-4.9) * t^2
Dividing both sides by -4.9:
t^2 = -80 / -4.9
t^2 ≈ 16.33
Taking the square root of both sides:
t ≈ √16.33
t ≈ 4.04 seconds
Therefore, it takes approximately 4.04 seconds to reach the ground.
1. V^2 = Vo^2 + 2g*h. Vo = 0, g = 9.8 m/s^2, h = 80 m, V = ?.
2. 0.5g*t^2 = 80. g = 9.8 m/s^2, t = ?.