The position of a ball as its kicked upwards is measured as
y(t) =−(4.9 t^2 + 20t + 0.50)m,
where t is measured in seconds. Find the vertical component of velocity and acceleration of the ball as a function of time. If the ground is located at yg = 0, find the time at which the ball hits the
ground.
Your equation is wrong with a minus sign parentheses outside the paretheses. Please verify that you copied it right.
It should be
y(t) = −4.9 t^2 + 20t + 0.50 (meters)
Zero vertical velocity is attained when
dy/dt = -9.8 t + 20 = 0
t = 20/9.8 = 2.04 seconds
It hits the ground when
−4.9 t^2 + 20t + 0.50 = 0
I suggest you use the quadratic equation and take the positive root.
You should get 4.016 s
To find the vertical component of velocity, we need to find the first derivative of the position equation with respect to time (t). This will give us the rate of change of position with respect to time, or the velocity.
Given the position equation: y(t) = -(4.9t^2 + 20t + 0.50)m
We can differentiate it with respect to time (t) using the power rule of differentiation:
dy/dt = -9.8t - 20
Hence, the vertical component of velocity is given by:
v(t) = dy/dt = -9.8t - 20 m/s
Moving on to finding the acceleration, we need to differentiate the velocity equation with respect to time (t) to get the rate of change of velocity with respect to time, or the acceleration.
Now, we differentiate the velocity equation:
dv(t)/dt = -9.8 m/s^2
Hence, the vertical component of acceleration is constant and given by:
a(t) = dv(t)/dt = -9.8 m/s^2
Now, to find the time at which the ball hits the ground, we need to set the position equation equal to zero and solve for t:
-4.9t^2 - 20t - 0.50 = 0
This is a quadratic equation in t. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -4.9, b = -20, and c = -0.50.
Using these values, we can substitute them into the quadratic formula:
t = [-(b) ± √((b)^2 - 4(a)(c))] / 2(a)
t = [-(20) ± √((-20)^2 - 4(-4.9)(-0.50))] / 2(-4.9)
Simplifying this expression will give us the solution for t, which represents the time at which the ball hits the ground.