Physics help please?

Question

Ok, I know these two are probably easy questions but something is throwing me off so please help. I'm not here so people can do my homework but I tried these two questions for too long and I'm getting a headache, I'm just not getting the right answer. Please help me in anyway; any help is GREATLY appreciated:

1. A ball is dropped from a cliff that is 150 metres above the ground. Two seconds later another ball is thrown downwards. Determine the velocity of the second ball if they both hit the ground at the same time.

2. A ball is thrown upwards with a velocity of 24.5 metres/second from the edge of a cliff that is 200 metres above the ground. Determine how much time can pass before another ball is dropped from the cliff so that each ball hits the ground at the same.

Answer 1

1. Given, h = 150 m
In case of the first ball, it is a freely falling body.
Therefore, u=0, a= +g

s = ut + 1/2 a t^2
i.e h = 1/2 g t^2
150 = 1/2 (10) t^2
150 = 5t^2
t^2 = 30
t = 5.48
Therefore, time taken by the freely falling body i.e 1st body to cover 150 m = 5.48 seconds.

Given, another ball is THROWN downwards with some velocity after 2 seconds. Therefore, time taken by this second body is equal to time taken by freely falling body minus 2
i.e t (time taken for second body) = 5.48 - 2
= 3.48

Now, for the second body, s = ut + 1/2 g t^2
150 = 3.48u + 1/2(10) (3.48)^2
150 = 3.48u + 5(12.1)
150 = 3.48u + 60.5
3.48u = 89.5
u = 89.5/3.48
u = 25.7 m/s

2. I'll give this answer in a moment.
Here, i've solved the problem:

For a body which is thrown vertically upwards from a height h, the formula is : h = -ut + 1/2 g t^2

For the ball thrown upwards:
h = -ut + 1/2g t^2
200 = -(24.5)t + 1/2(10)t^2
200 = -(24.5)t + 5 t^2
5t^2 -24.5t - 200 = 0
On solving the quadratic equation, we get t = 9.23 approx.

For the another ball that is dropped from the cliff, its a freely falling body.
s = ut + 1/2 a t ^2
h = 1/2g t^2 (Since u = 0 for a freely falling body)
200 = 1/2 (10) t^2
200 = 5 t^2
t^2 = 40
t = 6.32 approx.

The time difference between the two balls gives the time that can pass before the second ball is dropped so that both hit the ground at the same instant.
So, the answer is: (time taken by first body) - (time taken by dropped body)
i.e 9.23 - 6.32
= 2.91 seconds

Answer 2

1. A ball is dropped from a cliff that is 150 metres above the ground. Two seconds later another ball is thrown downwards. Determine the velocity of the second ball if they both hit the ground at the same time.

Find the time the first ball takes to hit the ground. Subtract 2 seconds from that. Then use the equation for distance with an initial velocity to solve for the velocity, given the time calculated.


2. A ball is thrown upwards with a velocity of 24.5 metres/second from the edge of a cliff that is 200 metres above the ground. Determine how much time can pass before another ball is dropped from the cliff so that each ball hits the ground at the same.

A similar problem, but in reverse. The thrown ball will have same velocity on its way down when it crosses the 200 m altitude. Calculate the total time it takes to go up and then hit the ground. Calculate the time required for the dropped ball to hit the ground.

Answer 3

Use the first ball's decent to determine the time. Then you have the time for the second ball's decent. The time and distance would be applied to the equation to give the original and end speed neglecting wind resistance or terminal velocity.

Find the amount of time it takes for the second ball to hit the ground. Then find the amount of time it takes for the first ball to go up and then down to the ground. Then subtract one from the other to get the time for the delay.

Similar to the other guy's answer. You get the idea.

Answer 4

do you use gravity as 9.8 or 10?

1. for 9.8
the second ball is thrown down at approximately 25.7m/s.
For 10
the second ball is thrown down at approximately 25.4m/s.
2.for 9.8
drop the second ball 3.1 seconds after the first
for 10
drop the second ball 2.9 seconds afteer the first