Calculate Velocity of Gravity: Ball Tossed Up 15 m/s from 12m

[BuhRock]

1. A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground.* Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).*

How high does it rise and how long does it take to get to its highest point?*

Homework Equations

3. I really don't know where to start. When it says assume a uniform downward acceleration of 10m/s^2, does that mean that for every second, the velocity goes down 10m/s? So after the first second, the velocity is now 5m/s?

 

[Uniquebum]

The basic equations of motion:
x = x_0 + v_0t + \frac{1}{2}*at^2
v = v_0 + at

Use these.

 

[BuhRock]

How can I use those if I don't have time?

I also found this equation:

s = sqrt(2(a)(12)+(15)^2) which gives me 21.5 meters. What is the 21.5 meters? Is that the highest point in the air?

 

[WatermelonPig]

That is the displacement, if you calculated it correctly. Add the original position to find the height.

For constant acceleration: vf^2 = v0^2 + 2ad.

And yeah, I need to learn LaTex.

 

[BuhRock]

So that would be adding 12 to 21.5, which is 33.5m. So this is the highest point the ball reaches?

 

[WatermelonPig]

d = (vf^2 - v0^2)/2a = (0^2 - 15^2)/2(-10) = 225/20 = 11.25m.
y = 11.25m + 12m = 23.25m.

 

[BuhRock]

Oh, I had 10, instead of -10. That's because it's a downward acceleration?

 

[Uniquebum]

v_0 = 15m/s, h_1 = 12m, v_{top} = 0
Lets choose down as the positive way of moving.
Thus
v = -v_0 + at ==> 0 = -v_0 + at ==> t = \frac{v_0}{a}
So the highest point will be
h = h_0 + v_0t + \frac{1}{2}at^2 ==> h = h_1 -v_0\frac{v_0}{a} + \frac{1}{2}a(\frac{v_0}{a})^2
Try the rest yourself.

 

[BuhRock]

Ok, so I got that. It takes 1.5 seconds to reach it's maximum height of 23.25 meters.

 

[HallsofIvy]

The object goes up as long as its velocity is positive. It is coming down once its velocity is negative. It is at its maximum height when its velocity is 0. You find the "t" for that by setting the velocity equal to 0 and solving for t.

 

[BuhRock]

The object goes up as long as its velocity is positive. It is coming down once its velocity is negative. It is at its maximum height when its velocity is 0. You find the "t" for that by setting the velocity equal to 0 and solving for t.

I understood everything up until you say set velocity equal to 0 and solve for t. If t = v/a Then how can you do 0/a? It will always be 0?

 

[gneill]

I understood everything up until you say set velocity equal to 0 and solve for t. If t = v/a Then how can you do 0/a? It will always be 0?

Strictly speaking, Δt = Δv/a. If you've got some initial velocity v_o and the final velocity is 0, then Δv is -v_o.

 

[BuhRock]

Strictly speaking, Δt = Δv/a. If you've got some initial velocity v_o and the final velocity is 0, then Δv is -v_o.

Oh I see, so what if it asks for this?

How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?*

I still never got an answer if when it says downward uniform acceleration of 10m/s^2 means that for every second, the avg velocity goes down by 10m/s?

 

[gneill]

Oh I see, so what if it asks for this?

How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?*

These questions can be answered by choosing appropriate kinematic equations and solving for the desired variable. (The list of general kinematic equations is quite small, and you should probably memorize them).

For the first case you can use either a conservation of energy approach, or the v_f² = v_i² + 2ad equation you've already seen. For the second case use the position vs time with initial conditions form of the equation of motion. Solve for time.

I still never got an answer if when it says downward uniform acceleration of 10m/s^2 means that for every second, the avg velocity goes down by 10m/s?

That's what the units of acceleration are interpreted to mean: Meters per second, per second