How Do You Solve Projectile Motion Problem #2?

[phys1618]

Please help! projectile motion problem #2

Homework Statement

A girl throws a rock straight out form the roof of a building. the rock has a velocity of 25 m/s when it leaves the girl's hand, and the rock hits the ground 3 seconds later.
1. how far from the bas e of the building does the rock land?
2. how tall is the building?
3. with what velocity does teh rock hit the ground?
4. how far from the top of the builidng does the rock land?

Homework Equations

d=v_it + (1/2)at²
c²= a² + b²
tan theta= opp/adj
a=(vf-vi)/t

The Attempt at a Solution

I'm unsure if my answers are correct. however, I tried and this is what i got
1. 75m
2. 45m
3. 39.05 m/s at 50.19^o below ground
4. 87.46m (this is the one I'm really unsure about)

all help and inputs are greatly appreciated. thank you for your time and efforts.

 

[LowlyPion]

Homework Statement

A girl throws a rock straight out form the roof of a building. the rock has a velocity of 25 m/s when it leaves the girl's hand, and the rock hits the ground 3 seconds later.
1. how far from the bas e of the building does the rock land?
2. how tall is the building?
3. with what velocity does teh rock hit the ground?
4. how far from the top of the builidng does the rock land?

Homework Equations

d=v_it + (1/2)at²
c²= a² + b²
tan theta= opp/adj
a=(vf-vi)/t

The Attempt at a Solution

I'm unsure if my answers are correct. however, I tried and this is what i got
1. 75m
2. 45m
3. 39.05 m/s at 50.19^o below ground
4. 87.46m (this is the one I'm really unsure about)

all help and inputs are greatly appreciated. thank you for your time and efforts.

For 2, if you used 10 as acceleration of gravity then that would be correct. But the more correct answer would use gravity as 9.8m/s²

I get a different answer for 3 by about 7 degrees but I think that may have to do with your treatment of part 2. I would guess your method is ok.
Part 4 is good enough for Government work.

Cheers.

 

[baba89]

Hello, I'm happy to assist you with this projectile motion problem. Firstly, I want to commend you for attempting to solve the problem on your own. Your answers seem to be in the right ballpark, but let's go through the problem together to make sure we get the correct solutions.

1. To solve for the horizontal distance the rock travels, we can use the equation d = vit, where d is the distance, vi is the initial velocity, and t is the time. In this case, we have d = (25 m/s)(3 s) = 75 m. So, the rock lands 75 meters away from the base of the building.

2. To find the height of the building, we can use the same equation d = vit, but this time the distance d represents the height of the building. So, we have d = (25 m/s)(3 s) = 75 m. This means that the building is 75 meters tall.

3. To find the velocity at which the rock hits the ground, we can use the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, and a is the acceleration due to gravity. In this case, we know that the final velocity is 0 m/s (since the rock hits the ground), the initial velocity is 25 m/s, and the acceleration due to gravity is -9.8 m/s^2 (since it is acting in the opposite direction of the initial velocity). So, we have 0 = 25 m/s + (-9.8 m/s^2)t. Solving for t, we get t = 2.55 s. Now, we can use this value of t in the equation vf = vi + at to find the final velocity: vf = 25 m/s + (-9.8 m/s^2)(2.55 s) = 0 m/s. So, the rock hits the ground with a velocity of 0 m/s.

4. This question is a bit tricky, but we can use the equation d = vit + (1/2)at^2 to solve for the vertical distance the rock travels. In this case, d represents the height of the building, vi is the initial velocity, a is the acceleration due to gravity, and t is the time. So, we have d = (25 m/s)(3 s) + (1/2)(