Homework questions, help or hints much appreciated =D

[moshilover]

1. A ball is thrown upward. Its initial vertical speed is 10.4 m/s, acceleration of gravity is 9.8 m/s^2, and maximum height hmax are shown in the figure below. What is its height, hmax? Answer in units of m. Neglect: air resistance

For this one I used 3 different formulas in different ways and each answer I got was wrong when I checked it. I divided 10.4 and 9.8 and got .9423076923 for time. Then I plugged that into a distance formula of d=ViT+1/2aT^2 and got 16.55510205 which was wrong.

2.A ball is thrown upward. Its initial vertical speed is v0, acceleration of gravity is 9.8 m/s^2, and maximum height is 5.5 m, as shown in the figure below. What is its initial vertical speed, v0? Answer in units of m/s. Neglect: Air resistance

So since the initial vertical speed is v0 wouldn't the initial velocity just be 0?? And so the answer is 0??

 

[radou]

1. A ball is thrown upward. Its initial vertical speed is 10.4 m/s, acceleration of gravity is 9.8 m/s^2, and maximum height hmax are shown in the figure below. What is its height, hmax? Answer in units of m. Neglect: air resistance

For this one I used 3 different formulas in different ways and each answer I got was wrong when I checked it. I divided 10.4 and 9.8 and got .9423076923 for time. Then I plugged that into a distance formula of d=ViT+1/2aT^2 and got 16.55510205 which was wrong.

Did you plug in the negative sign for gravitational acceleration?

2.A ball is thrown upward. Its initial vertical speed is v0, acceleration of gravity is 9.8 m/s^2, and maximum height is 5.5 m, as shown in the figure below. What is its initial vertical speed, v0? Answer in units of m/s. Neglect: Air resistance

What does the velocity of the ball equal at the point of the maximum height?

 

[m2003]

Hello,

Thank you for reaching out for help with your homework questions. I am happy to provide you with some guidance and hints to help you solve these problems.

1. When solving for the maximum height, hmax, of a ball thrown upward, we can use the formula hmax = (v0)^2/2g, where v0 is the initial vertical speed and g is the acceleration of gravity. We know that v0 = 10.4 m/s and g = 9.8 m/s^2. Plugging these values into the formula, we get hmax = (10.4)^2/2(9.8) = 5.42 m. Therefore, the maximum height of the ball is 5.42 m.

2. For the second problem, you are correct that the initial vertical speed, v0, is 0 since the ball is thrown upward with an initial vertical velocity of 0. However, we can also use the formula hmax = (v0)^2/2g to solve for v0. Rearranging the formula, we get v0 = √(2ghmax). Plugging in the given values, we get v0 = √(2(9.8)(5.5)) = 9.9 m/s. So, the initial vertical speed of the ball is 9.9 m/s.

I hope this helps guide you in the right direction to solve these problems. Remember to always double check your calculations and units to make sure your answer is correct. Good luck!