Solving the Last Problem: Finding the Maximum Height

[fisselt]

Last problem I can't get around..

Homework Statement

A ball is thrown upwards at some velocity. 2 seconds later the ball is caught 10m up on a building. What is the highest possible height of the ball?

Homework Equations

X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where X_f=10)
f(t)=v_x(t)-(9.8m/s^2)t^2 (set to 0)
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where t=1.5)

The Attempt at a Solution

I have X_f set to 10 and find the velocity of 14.8m/s. This means that the maximum height was 11.175m. This seems a little off as the ball should have dropped 0.05meters farther than the 1.175meters from my calculations. Am I making a mistake with the first equation?

 

[PeterO]

Last problem I can't get around..

Homework Statement

A ball is thrown upwards at some velocity. 2 seconds later the ball is caught 10m up on a building. What is the highest possible height of the ball?

Homework Equations

X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where X_f=10)
f(t)=v_x(t)-(9.8m/s^2)t^2 (set to 0)
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where t=1.5)

The Attempt at a Solution

I have X_f set to 10 and find the velocity of 14.8m/s. This means that the maximum height was 11.175m. This seems a little off as the ball should have dropped 0.05meters farther than the 1.175meters from my calculations. Am I making a mistake with the first equation?

I like this equation (red above) - but you have not fully utilised it, since you were told X_f=10 when t = 2, so it is a simple calculation

 

[fisselt]

Not sure what you mean. I find V_x to be 14.8m/s from this equation. Then, plug it into the second and set to 0 to find when the ball is stationary somewhere above 10m. Finally, I plug that time into the first equation again to find the height where V_x=0.

Maybe you can explain?

Thanks

 

[PeterO]

Not sure what you mean. I find V_x to be 14.8m/s from this equation. Then, plug it into the second and set to 0 to find when the ball is stationary somewhere above 10m. Finally, I plug that time into the first equation again to find the height where V_x=0.

Maybe you can explain?

Thanks

If you throw something up with a speed of 14.8 m/s, how high does it go?

 

[fisselt]

f(t)=v_x(t)-(9.8m/s^2)t^2 (set to 0)
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where t=1.5)

I found that after 1.5 sec the velocity and force of gravity were equal. So, plugging that into the second equation I got 11.175 meters.

Maybe my answer is correct, but it just seems a tiny bit off from the distance traveled by a ball falling that remaining .5 seconds. It should fall 1.225m but I have only 1.175m from the top of the balls height to the 10 meters.

 

[PeterO]

Last problem I can't get around..

Homework Statement

A ball is thrown upwards at some velocity. 2 seconds later the ball is caught 10m up on a building. What is the highest possible height of the ball?

Homework Equations

X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where X_f=10)
f(t)=v_x(t)-(9.8m/s^2)t^2 (set to 0)
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where t=1.5)

The Attempt at a Solution

I have X_f set to 10 and find the velocity of 14.8m/s. This means that the maximum height was 11.175m. This seems a little off as the ball should have dropped 0.05meters farther than the 1.175meters from my calculations. Am I making a mistake with the first equation?

On second reading of this, the only mystery I find is why you thought it should drop an extra 0.05 metres ?

 

[PeterO]

f(t)=v_x(t)-(9.8m/s^2)t^2 (set to 0)
X_f=x_i+V_x(t)-(1/2)(9.8m/s^2)t^2 (where t=1.5)

I found that after 1.5 sec the velocity and force of gravity were equal. So, plugging that into the second equation I got 11.175 meters.

Maybe my answer is correct, but it just seems a tiny bit off from the distance traveled by a ball falling that remaining .5 seconds. It should fall 1.225m but I have only 1.175m from the top of the balls height to the 10 meters.

Strange statement [red above]. That is like saying while riding my bicycle I found my speed and the reaction force from the saddle were equal?

EDIT: perhaps the idea that the ball was at maximum height after exactly 1.5 seconds was wrong?

 

[fisselt]

I shouldn't have said equal but rather after 1.5 seconds the balls velocity succumbed to gravity and had 0 velocity.

Shoot, I figured my problem. I wanted to check my answer so I used .5 seconds of freefall equals 1.225m. Which I believe is true but the time wasn't exactly 1.5seconds. I didn't round it for my final calculation but for some reason I did for checking it. The ball is actually falling closer to 0.49seconds which is very close to 1.175m.

Thanks for the help.. sometimes just sitting down and thinking how to explain the problem helps figure it out!