A ball is thrown upward with an initial velocity of 64 feet per second

Question

A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. How long does it take for the ball to hit the ground? (Acceleration due to gravity is a constant of -32 feet per second per second.)

Answer 1

Hi Sofia,

We can solve this by using a memorized kinematic equation, but since you tagged this as a Calculus question, we can derive this equation from intuition:

let's begin with the simple observation that all objects close to the surface of the Earth accelerate downwards with a constant acceleration of g. We can now write an acceleration equation for all such free falling objects. Let's also remember that our kinematic equations should be functions of time.

a(t) = - g

Now, we can also describe the velocity of such free falling objects by realizing that acceleration is the time derivative of velocity ( a = dv/dt ). So if we integrate our acceleration function with respect to time, we will end up with a velocity function:

v(t) = ∫ a dt = -gt + C (remember, this is an indefinite integral)

Now, what does C equal to in the above equation? If we set time equal to zero (that is, at the very beginning of our motion in this example), we get:

v(0) = -g(0) + C = C

This means that C must be equal to our initial velocity (velocity at time t = 0)

The equation becomes:

v(t) = - gt + v₀

Then, to get to a description of position, we can remember that velocity is the time derivative of position ( v = dx/dt ). So if we integrate our velocity function with respect to time, we will end up with a position function:

x(t) = ∫ v dt = -(1/2)gt² + v₀t + D (where D is just another integration constant like C was previously)

What does D equal to above? Let's set time equal to zero again:

x(0) = -(1/2)g(0)² + v₀(0) + D = D

This means that D must be equal to our initial position (x at t = 0)

So our final position equation becomes:

x(t) = -(1/2)gt² + v₀t + x₀

plugging in our constants for this scenario:

x(t) = -16t² + 64t + 80

When the object hits the ground, it's final x position (at t = t_f) will be zero, so:

x(t_f) = 0 = -16t_f² + 64t_f + 80

we can solve this by factoring (you can do this on your own) to get 2 solutions for t

t = -1 seconds (ignore this because negative time has no real meaning in our scenario)

t = 5 seconds (the answer)