How Do You Calculate the Time and Height of a Ball Thrown Upwards?

[calisoca]

Homework Statement

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?

Homework Equations

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

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 metersPart 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/sConclusion:

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.

 

[LowlyPion]

Looks OK.

Maybe you aren't so confused as you think?

 

[calisoca]

Well, here is where my confusion is.

Generally, you can express the equations for the ball as it is tossed upwards as:
a(t) = -9.8 m/(s^2)
v(t) = -9.8t + C
s(t) = -4.9(t^2) + Cx + D

Generally, you can express the equations for the ball as it falls as:
a(t) = -9.8 m/(s^t)
v(t) = -9.8t - C
s(t) = -4.9(t^2) - Cx + D

I am confused as to the correctness of the generally expressed equations for v(t) and s(t).

For v(t), my logic is that since the ball is now falling, instead of v(t) = -9.8t + C, the equation becomes v(t) = -9.8t - C. The C, which would be any initial velocity, is now negative, since the ball is falling. Is that logic correct?

For s(t), I'm not really sure why the D is not negative. In other words, I am unsure as to why the equation for s(t) when the ball is falling is not s(t) = -4.9(t^2) - Cx - D?

 

[LowlyPion]

It's all a matter of which direction you choose as positive. Gravity is down, so if you choose up as positive then g is (-) in your equations.

Once you've chosen your positive direction then you merely express the equations - with the applicable variables s,v,g - with their signs accordingly.

Hence you can even find the total time from launch upward to when it strikes the ground below (Y = 0) from just the one equation.

Y = Yo + Vo*t -1/2*g*t²

0 = 50 + 10*t - 4.9*t²

This quadratic yields 2 solutions, but 1 is negative and perhaps would apply if the ball had been thrown that much earlier from the ground. But the positive solution - the right side of time's arrow - is how long it takes to go up and then down.

As long as you are careful with your signs in constructing your equations you should be fine.

 

[calisoca]

LowlyPion, thank you so very much for your help. You have really helped me to understand what I was doing!