# Physics Questions on Thrown Up Ball & Bullet

• A.I. BOT
In summary, the first question asks about the maximum height reached by a ball thrown up with a speed of 30 m/s, and the correct answer is b) 46m. The second question asks about the initial speed of a bullet shot straight up and returning to its starting point in 10 seconds, with the correct answer being d) 98 m/s. The formula used to find the initial velocity in the second question is incorrect, as there should not be a 1/2 in the equation. The correct formula is Vi = Vf - at. The statement about the speed being the same when the object returns to its starting point is also incorrect, as the speed would be zero only if the object did not go up at
A.I. BOT
[Solved] Thrown up ...

I am stupid when it comes to Physics so please guide me :). I have 2 questions on a worksheet I got today and I am stuck.

A ball is thrown up with a speed of 30 m/s. What is the maximum height reached by the ball?
a) 23m b) 46m c) 92m d) 132m
Is it b? .. (30 m/s) / 9.8 m/s^2 = 3.06s ... 30 x 1/2(3.06) = 46m?

A bullet is shot stright up and returns to its starting point in 10 s. What is the initial speed of the bullet
a) 9.8 m/s b) 25 m/s c) 49 m/s d) 98 m/s
Is it b? .. Vi = Vf - a*1/2(t) ... Vi = 0, Vf = 0, a = 9.8, t = 1/2(10) = 5 ... Vi = 0 - 9.8*1/2(10) ... 25m?

Thanks :)

Last edited:
The first answer looks to be correct (not sure what the 2nd part is you used to find it), but I get something different on your second question.

"Vi = Vf - a*1/2(t)"

Are you sure there's a 1/2 there? http://en.wikipedia.org/wiki/Kinematic

Hmm your right, so it should be d) 98 m/s I am assuming ?

A.I. BOT said:
Hmm your right, so it should be d) 98 m/s I am assuming ?
How does the final velocity at the end of 10 seconds compare to the initial velocity?

I was told in school that when something is thrown up ... the speed it went up, will be the same speed coming down. The bullet is going to return to it's start point it states, thus making its final velocity 0m/s ? so ... If I just plug the numbers into Vi = Vf - at I get 98m/s, so I will go with that and see what it says when I get the assignment back.

Thanks for the help guys :)

A.I. BOT said:
I was told in school that when something is thrown up ... the speed it went up, will be the same speed coming down. The bullet is going to return to it's start point it states, thus making its final velocity 0m/s ? so ... If I just plug the numbers into Vi = Vf - at I get 98m/s, so I will go with that and see what it says when I get the assignment back.

Thanks for the help guys :)
Your statement is contradictory. The speed is the same when it gets back to the starting point that it had when thrown. That is not zero. If it were zero, it would never have gone up.

## 1. What factors affect the trajectory of a thrown up ball?

The trajectory of a thrown up ball is affected by the initial velocity, angle of release, air resistance, and gravity. These factors determine the path the ball will follow as it travels through the air.

## 2. How does the weight of a ball affect its trajectory when thrown up?

The weight of a ball does not significantly affect its trajectory when thrown up. The mass of the ball impacts its acceleration due to gravity, but it does not affect the angle or shape of its trajectory.

## 3. What is the difference between a ball thrown up with no initial velocity and one thrown up with an initial velocity?

A ball thrown up with no initial velocity will only travel the distance of its vertical displacement, while a ball thrown up with an initial velocity will follow a parabolic trajectory and travel further horizontally before falling back down.

## 4. How does air resistance impact the trajectory of a thrown up ball?

Air resistance, also known as drag, can cause a thrown up ball to experience a decrease in velocity and a change in its trajectory. This is because the force of air resistance acts opposite to the direction of motion and can slow the ball down.

## 5. Why does a bullet have a flatter trajectory than a thrown up ball?

A bullet has a flatter trajectory than a thrown up ball because it has a much higher initial velocity and less air resistance acting on it. This allows the bullet to maintain a more horizontal path before it is affected by gravity and falls back down.

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