Can you please check if my answers are correct?

[honestliar]

Homework Statement

An object starts from rest w/ a constant acceleration of 8m/s^2 along straight line.
a)find the speed at the end of 5 sec.------40sec
b)the ave. spped for the sec interval------20m/s

A stone is thrown vertically upward w/ an initial velocity of 49m/s
a)How long does it take to reach the highest point-----5 sec
b)How far does the stone go---122.5m
c)How fast does the stone go----4.52s



A stone is dropped from a bridge w/c is 60m above seal level. a)find the time before it hits the water----12.24s
b)Speed upon hitting the water---34.29m/s


thanks

 

[tiny-tim]

Hi honestliar!

They're ok except for …

a)find the time before it hits the water----12.24ss

(and I don't understand the question "c)How fast does the stone go" )

 

[kerimek]

Hello! I am not able to check your answers without more information and context. However, I can provide guidance on how to solve these problems.

For the first problem, we can use the equation v = u + at to find the speed at the end of 5 seconds. Here, v represents the final speed, u represents the initial speed (which is 0 since the object starts from rest), a represents the acceleration, and t represents time. Plugging in the values, we get v = 0 + (8m/s^2)(5s) = 40m/s. Therefore, your answer for part a) is correct.

For part b), we need to find the average speed for the given time interval. We can use the equation v = d/t, where v represents the average speed, d represents the distance, and t represents time. Since the object starts from rest, the initial speed is 0, so we can use the final speed (40m/s) as the average speed. Therefore, the average speed for the given time interval is 20m/s. Your answer is correct.

Moving on to the second problem, we can use the equation v = u + at to find the time it takes for the stone to reach the highest point. Here, v represents the final speed (which is 0 when the stone reaches the highest point), u represents the initial speed (49m/s), a represents the acceleration (due to gravity, it is -9.8m/s^2 since the stone is thrown vertically upward), and t represents time. Plugging in the values, we get 0 = 49m/s + (-9.8m/s^2)t. Solving for t, we get t = 5s. Therefore, your answer for part a) is correct.

For part b), we can use the equation s = ut + (1/2)at^2 to find the distance the stone goes. Here, s represents distance, u represents initial speed (49m/s), a represents acceleration (-9.8m/s^2), and t represents time. Plugging in the values, we get s = (49m/s)(5s) + (1/2)(-9.8m/s^2)(5s)^2 = 122.5m. Your answer is correct.

Lastly, for part c), we can use the same equation v =