How Do I Solve These Physics 1 Homework Problems?

[rkw6]

hey, I am taking physics 1 and i have homework, but I am not sure where to start or how to do the problems! my book is NOT helpful, so if anyone can help me it will be greatly appreiciated!


1)A ball is thrown upward from the ground with an initial speed of 23.3 m/s; at the same instant, a ball is dropped from a building 13.4 m high. After how long will the balls be at the same height?




2)A model rocket is launched straight upward with an initial speed of 49.6 m/s. It accelerates with a constant upward acceleration of 2.01 m/s2 until its engines stop at an altitude of 170 m. What is the maximum height reached by the rocket?How long after lift off does the rocket reach its maximum height? How long is the rocket in the air?




3)A startled armadillo leaps upward rising 0.444 m in 0.230 s. What was its initial speed?What is its speed at this height?How much higher does it go?





4)A particle moves along the x-axis according to the equation
x = 47 t + 8t2

(where t is in seconds and x is in meters). Calculate the average velocity of the particle during the first 3.0 s of its motion.Calculate the instantaneous velocity of the particle at t = 3.0 s.Calculate the instantaneous acceleration of the particle at t = 3.0 s.

If anyone can help with ANY of these, it would be a huge help!

 

[LowlyPion]

hey, I am taking physics 1 and i have homework, but I am not sure where to start or how to do the problems! my book is NOT helpful, so if anyone can help me it will be greatly appreiciated!

1)A ball is thrown upward from the ground with an initial speed of 23.3 m/s; at the same instant, a ball is dropped from a building 13.4 m high. After how long will the balls be at the same height?

2)A model rocket is launched straight upward with an initial speed of 49.6 m/s. It accelerates with a constant upward acceleration of 2.01 m/s2 until its engines stop at an altitude of 170 m. What is the maximum height reached by the rocket?How long after lift off does the rocket reach its maximum height? How long is the rocket in the air?

3)A startled armadillo leaps upward rising 0.444 m in 0.230 s. What was its initial speed?What is its speed at this height?How much higher does it go?

4)A particle moves along the x-axis according to the equation
x = 47 t + 8t2

(where t is in seconds and x is in meters). Calculate the average velocity of the particle during the first 3.0 s of its motion.Calculate the instantaneous velocity of the particle at t = 3.0 s.Calculate the instantaneous acceleration of the particle at t = 3.0 s.

If anyone can help with ANY of these, it would be a huge help!

Welcome to PF.

I'm sure your book has formulas that cover the material. Which ones do you think applies? People will help you sort it out if you have made an effort, but alas doing your work for you is not in the cards.

 

[baba89]

1) To solve this problem, we can use the equations of motion for constant acceleration. First, we need to set up equations for the vertical motion of each ball. For the ball thrown upward, we have: y = y0 + v0t - (1/2)gt^2, where y is the height, y0 is the initial height (which is 0 in this case), v0 is the initial velocity (23.3 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time. For the ball dropped from the building, we have: y = y0 - (1/2)gt^2, where y0 is the initial height (13.4 m) and all other variables are the same.

To find the time when the balls are at the same height, we can set the two equations equal to each other and solve for t. This gives us t = 2.46 seconds, meaning that after 2.46 seconds, both balls will be at the same height.

2) To solve this problem, we can use the equations of motion again. This time, we need to find the maximum height reached by the rocket, which is when its final velocity is 0. We can use the equation vf^2 = v0^2 + 2ad, where vf is the final velocity (0 m/s), v0 is the initial velocity (49.6 m/s), a is the acceleration (2.01 m/s^2), and d is the displacement (maximum height) of the rocket. Solving for d, we get a maximum height of 385.3 m.

To find the time when the rocket reaches its maximum height, we can use the equation vf = v0 + at, where vf is the final velocity (0 m/s), v0 is the initial velocity (49.6 m/s), a is the acceleration (2.01 m/s^2), and t is the time. Solving for t, we get a time of 24.6 seconds.

To find the total time the rocket is in the air, we can use the equation vf = v0 + at, where vf is the final velocity (0 m/s), v0 is the initial velocity (49.6 m/s), a is the acceleration (2.01 m/s^2), and t is the total