aj uses a lingshot to launch a rock straight up from the a point of 6 ft. above ground level with an initial velocity of 170 ft per second.
A) find an equation that models the height of the rock t seconds after it is launched.
b) what is the maximum height of the rock and when will it reach it. explain how you found this answer or show work.
c)when will the rock hit the ground? explain how you found this answer or show work.
d) what is the rock's vertical velocity when it hits the ground? show work
The initial velocity is in the y-direction.
A) It's height above ground is
y = x0 + v0*t - 1/2*g*t^2
where y is its height, x0 is its initial position, v0 is it's initial speed in the y direction, g is the acceleration due to gravity and t is time. Plugging in the values from the problem:
y = 6 + 170*t -1/2*g*t^2
b) The maximum height will be found at the point when dy/dt equals zero
dy/dt = 170 - 1/2*g*t^2 = 0
Solve for t, then plug this value back into y to find the maximum height
c) The rock will hit the ground when y = 0
0 = 6 + 170*t -1/2*g*t^2
Solve for t using the quadratic equation solutions
d) the velocity is dy/dt = 170 - 1/2*g*t^2
Plug your answer for t from part c into this equation to find the velocity when it hits the ground
To find the equations and solve the problem, we can use the basic equations of motion. These equations are derived from Newton's laws of motion.
a) To model the height of the rock, we can use the equation of motion for vertical motion:
h(t) = -16t^2 + vt + h0
Where:
h(t) represents the height of the rock at time t.
v represents the initial velocity of the rock.
h0 represents the initial height of the rock.
In this case, v = 170 ft/s (initial velocity), and h0 = 6 ft (initial height). Thus, the equation becomes:
h(t) = -16t^2 + 170t + 6
b) To find the maximum height of the rock and when it reaches it, we can use calculus. The maximum height occurs at the vertex of the parabolic trajectory described by the equation.
The vertex of a parabola with equation ax^2 + bx + c can be found using the formula: x = -b/(2a)
In this case, a = -16, b = 170, and c = 6. Calculating the vertex's x-coordinate:
t_vertex = -170 / (2 * -16)
t_vertex = 5.3125 seconds
To find the maximum height, plug the vertex's x-coordinate into the equation:
h_max = -16(5.3125)^2 + 170(5.3125) + 6
h_max ≈ 1411.375 ft
Therefore, the maximum height of the rock is approximately 1411.375 feet, reached at around 5.3125 seconds.
c) To determine when the rock hits the ground, we need to find the time when the height becomes zero. Set h(t) = 0 and solve for t:
-16t^2 + 170t + 6 = 0
This equation is a quadratic equation that can be solved using the quadratic formula. The solutions for t will give us the times when the rock hits the ground.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting a = -16, b = 170, and c = 6 into the formula:
t1 = (-170 + √(170^2 - 4 * -16 * 6)) / (2 * -16)
t2 = (-170 - √(170^2 - 4 * -16 * 6)) / (2 * -16)
t1 ≈ 11.453 seconds
t2 ≈ 0.148 seconds
Since time cannot be negative, the rock hits the ground at approximately 11.453 seconds.
d) To find the rock's vertical velocity when it hits the ground, we can calculate the derivative of the height equation with respect to time (t):
h'(t) = -32t + 170
At the time the rock hits the ground (t = 11.453 seconds):
h'(11.453) = -32(11.453) + 170
h'(11.453) ≈ -365.656 ft/s
Therefore, the rock's vertical velocity when it hits the ground is approximately -365.656 ft/s. The negative sign indicates that the velocity is directed downwards.