(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am really confused here. I think I understand the concepts, but my implementation of those concepts is all wrong.

The problem:

A ball is tossed upwards from an initial height of 50 meters, and it has an initial velocity of 10 meters per second.

1.) How long in seconds does it take the ball to reach its maximum height, and what is the maximum height?

2.) How far does the ball fall from its maximum height to hit the ground, how long in seconds does it take the ball to hit the ground, and what is the velocity of the ball when it hits the ground?

2. Relevant equations

a(t) = -9.8 m/(s^2)

3. The attempt at a solution

Part 1:

a(t) = -9.8 m/(s^2)

dv/dt = a(t)

v(t) = -9.8t + 10

ds/dt = v(t)

s(t) = -4.9(t^2) + 10t + 50

At v(t) = 0 m/s, the ball has reached its maximum height.

v(t) = -9.8t + 10

0 = -9.8t + 10

-10 = -9.8t

t = 1.020408 seconds

s(t) = -4.9(t^2) + 10t + 50

s(t) = -4.9(1.020408^2) + 10(1.020408) + 50

s(t) = 55.102 meters

Part 2:

Once the ball has reached its maximum height,

a(t) = -9.8 m/(s^2)

v(t) = 0 m/s

s(t) = 55.102 meters

dv/dt = a(t)

v(t) = -9.8t - 0 (any initial velocity would be negative since the ball is now falling)

OR SIMPLY v(t) = -9.8t

ds/dt = v(t)

s(t) = -4.9(t^2) - 0t + 55.102

OR SIMPLY s(t) = -4.9(t^2) + 55.102

At s(t) = 0, the ball has reached the bottom.

s(t) = -4.9(t^2) + 55.102

0 = -4.9(t^2) + 55.102

-55.102 = -4.9(t^2)

t = 3.353402 seconds

v(t) = -9.8(3.353402)

v(t) = -32.86 m/s

Conclusion:

Thus, when tossed upwards from a height of 50 meters at an initial velocity of 10 meters per second, the ball travels 5.102 meters upwards in 1.020408 seconds to reach its peak at 55.102 meters.

The ball then falls 55.102 meters down to the ground in 3.353402 seconds, and it hits the ground with a final velocity of 32.86 meters per second.

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# Homework Help: This is easy, I'm just highly confused.

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