How Does Projectile Motion Affect a Ball Thrown Upward from a Building?

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Homework Help Overview

The discussion revolves around a problem involving projectile motion, specifically analyzing the motion of a ball thrown vertically upward from a building. The problem includes calculating various parameters such as position, velocity, maximum height, and acceleration at different points in time.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for position and velocity at specific times, questioning the accuracy of their results. There is confusion regarding the time to reach a height of 5 m, with different answers being proposed. Some participants seek clarification on the equations used and the assumptions made in their calculations.

Discussion Status

The discussion is active, with participants providing feedback on each other's calculations and questioning the setup of certain equations. Some guidance has been offered regarding the correct formulation of the equations, but there is no explicit consensus on the answers to all parts of the problem.

Contextual Notes

Participants express uncertainty about their calculations and the equations used, particularly in relation to the time to reach a specific height. There is also mention of homework constraints that may affect their approach to the problem.

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Homework Statement


You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; the ball is then in free fall. On its way back down, it just misses the railing.
Find (a) the position and velocity of the ball 1.0 s and 4.0 s after leaving your hand;
(b) the velocity when the ball is 5.0 m above the railing;
(b.1) How long will the ball reach 5 m
(c) the maximum height reached and the time at which it is reached; and
(d) the acceleration of the ball when it is at its maximum height.

Homework Equations


Freefall equation like Vf=Vo-gt and etc

The Attempt at a Solution


T at 1
a.) ΔY = (15 m/s)(1 s) - (1/2)(9.8 m/s^2)(1 s)^2
ΔY = 10.1 m

T at 4
ΔY = -18.4 m

V at 1
Vf = (15 m/s) - (9.8 m/s^2)(1)
= 5.2 m/s

V at 4
Vf = -24.2 m/s

b.) Vf^2 = Vi^2 - 2gΔy
Vf^2 = (15 m/s)^2 - (2)(9.8 m/s^2)(5 m)
Vf^2 = 11.27 m/s

b.1) (11.27 m/s) = (15 m/s) - (9.8^2)(t)
t = 0.38

5 m = (15 m/s)(t) - (1/2)(9.8 m/s^2)(t)
t = 2.68

I'm a bit confuse here I got 2 different answer in this part. Really need a big help here

c.) a = Δv/Δt -> t = Δv/a -> t = (15 m/s)/(9.8 m/s^2)
tmax = 1.5306

Δymax = (15 m/s)(1.5306 s) - (1/2)(9.8 m/s^2)(1.5306 s)^2
y = 11.48

d.) The answer is -9.8 m/s^2 right? Constant?
 
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For question b.1) you wrote 5=15t-9,8*0.5t while it should be 5=15t-9.8*0.5T^2
 
Sobhan said:
For question b.1) you wrote 5=15t-9,8*0.5t while it should be 5=15t-9.8*0.5T^2

lol I am such a stupid one

is my c.) and d.) right?
 
VaioZ said:
lol I am such a stupid one

is my c.) and d.) right?
Yes, c and d are right.
 

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