Near the surface of the earth, the acceleration of a falling body due to gravity is 32 feet per second per second, provided that air resistance is neglected. If an object is thrown upward from and initial height of 3000 feet with a velocity of 75 feet per second, find its velocity and height 6 seconds later.
Velocity =
Height =
it is well known that the equations are
s(t) = 3000+75t-16t^2
v(t) = 75-32t
So plug and chug
Extra credit: derive those equations.
hint
if h = height
d^2h/dt^2 = acceleration = -32 ft/s^2 constant
dh/dt = speed up
To find the velocity and height of the object 6 seconds later, we can use the equations of motion for freely falling objects.
1. Velocity equation: v = u + at
2. Height equation: h = u*t + (1/2)*a*t^2
Given:
Initial velocity, u = 75 feet per second (upward)
Acceleration, a = -32 feet per second squared (downward, considering the direction)
To find the velocity after 6 seconds:
1. Using the velocity equation:
v = u + at
v = 75 + (-32)*6
v = 75 - 192
v = -117 feet per second (negative sign indicates the direction is downward)
Therefore, the velocity of the object 6 seconds later is -117 feet per second.
To find the height after 6 seconds:
2. Using the height equation:
h = u*t + (1/2)*a*t^2
h = 75*6 + (1/2)*(-32)*(6)^2
h = 450 - 576
h = -126 feet (negative sign indicates below the initial height)
Therefore, the height of the object 6 seconds later is -126 feet below the initial height (3000 feet).
To find the velocity and height of the object 6 seconds later, we can use the equations of motion.
The equation for velocity is given by:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Plugging in the values:
u = 75 feet per second (initial velocity)
a = -32 feet per second squared (acceleration due to gravity, taking negative sign as the object is thrown upward)
t = 6 seconds
Applying the equation, we have:
v = 75 + (-32) * 6
v = 75 - 192
v = -117 feet per second (negative sign indicates that the object is moving upwards)
So, the velocity 6 seconds later is -117 feet per second.
Now, to find the height, we can use the equation:
s = ut + (1/2)at^2
Where:
s = height
u = initial velocity
t = time
a = acceleration
Plugging in the values:
u = 75 feet per second (initial velocity)
a = -32 feet per second squared (acceleration due to gravity, taking negative sign as the object is thrown upward)
t = 6 seconds
Applying the equation, we have:
s = 75 * 6 + (1/2) * (-32) * (6^2)
s = 450 + (-16) * 36
s = 450 - 576
s = -126 feet (negative sign indicates that the object is above the initial height of 3000 feet)
So, the height 6 seconds later is -126 feet.