Asked by justin
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.
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.
Answers
Answered by
drwls
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
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
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