The height h after t seconds of a falling object starting at t=0 at height x feet is h(t)=x−16t^2 (feet).
Let t_1 be the time it takes the object to fall x feet. The kinetic energy of a ball of mass m dropped vertically x feet is E_1=(1/2)m(v_1)^2 , where v_1=h′(t_1). Find the formula for E_1 in terms of m and x.
I've tried substituting values to only get in terms of m and x but am having trouble. Please help, thank you.
To find the formula for E_1 in terms of m and x, we first need to find the derivative of the height function h(t) with respect to time t, which will give us the velocity function.
Given h(t) = x - 16t^2, we can find h'(t), the derivative of h(t), using the power rule for differentiation. Thus,
h'(t) = d/dt (x - 16t^2) = 0 - 32t = -32t.
Now, we need to determine the time it takes for the object to fall x feet, denoted as t_1. Since the object is at height x when t = t_1, we can set h(t_1) = x and solve for t_1.
x - 16t_1^2 = x
-16t_1^2 = 0
t_1^2 = 0
t_1 = 0
Therefore, t_1 = 0.
Now, we can find v_1, which is h'(t_1).
v_1 = h'(0) = -32(0) = 0.
Substituting v_1 = h'(t_1) = 0 into the formula for kinetic energy, E_1 = (1/2)m(v_1)^2, we get:
E_1 = (1/2)m(0)^2
E_1 = (1/2)m(0)
E_1 = 0.
Therefore, the formula for E_1 in terms of m and x is E_1 = 0.
In other words, the kinetic energy of a ball dropped vertically from a height x feet is zero. This may seem counterintuitive, but it is because at the moment the ball reaches the height x, its velocity is momentarily zero, resulting in zero kinetic energy.