A stone is thrown straight downward with initial speed 8.0m/s from height of 20m find (a) the time it takes to reach the ground and (b) speed with which it strikes.
Force = - m g = -m * 9.81 m/s^2 = m a
so
a = -9.81 m/s^2
v = Vi + a t = -8.0 - 9.81 t
h = Hi + Vi t + (1/2) a t^2 = 20 - 8.0 t - 4.9 t^2
ground when h = 0
4.9 t^2 + 8.0 t - 20 = 0
t = 1.36 seconds is positive time root of quadratic
v = -8.0 - 9.81 (1.36) = -21.3 meters/seconds
speed is absolute value of velocity = 21.3 m/s
ALTERNATELY
energy at top = m g Hi + (1/2) m Vi^2 = m(9.81*20+.5*64)
= 228 * m Joules
energy at bottom = (1/2) m v^2
so
m v^2 = 2 * 228 * m
v^2 = 456
v = 21.3 m/s (again)
average speed = .5 (8+21.3) = 14.7 m/s
distance = 20
so time = 1.36 seconds (again)
To find the time it takes for the stone to reach the ground, we can use the kinematic equation:
h = ut + (1/2)gt^2
Where:
h = initial height = 20m
u = initial velocity = 8.0m/s (negative sign since it is thrown downward)
g = acceleration due to gravity = 9.8m/s^2
t = time
(a) Solving for time:
20 = (-8.0)t + (1/2)(9.8)t^2
To solve this quadratic equation, we bring it to the standard form:
0 = (1/2)(9.8)t^2 - 8.0t + 20
Now, we can solve for t. There are a few ways to solve a quadratic equation. One straightforward way is to use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = (1/2)(9.8), b = -8.0, and c = 20. Plugging in these values:
t = [-(8.0) ± √((-8.0)^2 - 4(1/2)(9.8)(20))] / [2(1/2)(9.8)]
Simplifying this equation will give us two possible values for time, but since we are dealing with a physical situation, we need to choose the positive value for time.
(b) To find the speed with which the stone strikes the ground, we can use the equation:
v = u + gt
Where:
v = final velocity (speed with which it strikes)
u = initial velocity
g = acceleration due to gravity
t = time
We have the values for u and t, so we can calculate the final velocity.