Question

An object is thrown upward with an initial speed of 18.5 m/s from a location 12.2 m above the ground. After reaching its maximum height it falls to the ground. What will its speed be in the last instant prior to its striking the ground?

Answer 1

Answer:

The speed of the object in the last instant prior to hitting the ground is -24.1 m/s

Explanation:

The equation for the position and velocity of the object will be:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the object at time t

v0 = initial velocity

y0 = initial height

g = acceleration due to gravity

t = time

v = velocity at time t

We know that at its maximum height, the velocity of the object is 0. We can obtain the time it takes the object to reach the maximum height and with that time we can calculate the maximum height:

v = v0 + g · t

0 = 18.5 m/s - 9.8 m/s² · t

-18.5 m/s / -9.8 m/s² = t

t = 1.89 s

Now,let´s find the max-height:

y = y0 + v0 · t + 1/2 · g · t²

y = 12.2 m + 18.5 m/s · 1.89 s + 1/2 ·(-9.8 m/s²) · (1.89 s)²

y = 29.7 m

Now, let´s see how much it takes the object to hit the ground:

In that instant, y = 0.

y = y0 + v0 · t + 1/2 · g · t²

0 = 29.7 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²   (notice that v0 = 0 because the object starts from its maximum height, where v = 0)

-29.7 m = -4.9 m/s² · t²

t² = -29.7 m / -4.9 m/s²

t = 2.46 s

Now, we can calculate the speed at t= 2.46 s, the instant prior to hitting the ground.

v = v0 + g · t

v = g · t

v = -9.8 m/s² · 2.46 s

v = -24.1 m/s