I am standing on the roof of a building with a pea shooter, and I shoot a pea up into the air from a height of 80 feet. The pea reaches its maximum height 2 seconds after I shoot, then falls to the ground.
Give the position of the pea as a function of time
in foot system g = -32 ft/s^2
Hi = 80
v = Vi - 32 t
at max height v = 0
so Vi = 32 t at top
so Vi = 32 * 2 = 64 ft/s
h = Hi + Vi t - 16 t^2
h = 80 + 64 t - 16 t^2
h = 80 + 16 t^2 at top when
To find the position of the pea as a function of time, we need to consider the motion of the pea in two separate phases: the upward motion and the downward motion.
During the upward motion, the pea moves against gravity, slowing down until it reaches its maximum height. During this phase, we can use the equation for vertical motion:
h(t) = h0 + v0*t - (1/2)*g*t^2
Where:
- h(t) is the height of the pea at time t
- h0 is the initial height (80 feet in this case)
- v0 is the initial velocity (which we need to find)
- g is the acceleration due to gravity (approximately -32.2 feet/second^2)
Since the pea reaches its maximum height after 2 seconds and then starts falling, we know that its velocity at the top of its trajectory is 0. Therefore, we can calculate v0 using the equation:
v(t) = v0 - g*t
When t = 2 seconds, v(t) = 0. Solving for v0:
0 = v0 - g*2
v0 = 2g
Substituting this value of v0 into the equation for vertical motion, we get:
h(t) = h0 + 2gt - (1/2)g*t^2
Now, during the downward motion, the pea falls freely under gravity. We can still use the equation for vertical motion, but with a negative gravitational acceleration:
h(t) = h1 + v1*t - (1/2)*(-g)*t^2
Where:
- h(t) is the height of the pea at time t during the fall
- h1 is the maximum height reached by the pea (which we need to find)
- v1 is the velocity of the pea as it starts falling (which we can calculate)
Since the pea reaches its maximum height at t = 2 seconds, we substitute t = 2 into the equation for the upward motion and solve for h1:
h1 = h(2) = h0 + 2g*2 - (1/2)g*(2^2)
h1 = h0 + 4g - 2g
h1 = h0 + 2g
Now, to find v1, we differentiate the equation for h(t) with respect to time:
v(t) = v1 - (-g)*t
v1 = -g*t + v(t)
At t = 2 seconds, v(t) = 0. Therefore,
v1 = -g*2 + 0
v1 = -2g
Substituting these values into the equation for the downward motion, we get:
h(t) = h1 - 2g*t - (1/2)(-g)*t^2
= (h0 + 2g) - 2g*t + (1/2)g*t^2
So, the position of the pea as a function of time is given by:
h(t) = h(t) =
{ h0 + 2gt - (1/2)g*t^2 for 0 <= t <= 2
{ (h0 + 2g) - 2g*t + (1/2)g*t^2 for t > 2
This equation describes the pea's height at any given time after it is shot into the air.