How Long to Reach Peak Height for a Thrown Object?

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Homework Help Overview

The discussion revolves around determining the time required for an object thrown upwards with an initial vertical velocity of 15 m/s and a horizontal velocity of 18 m/s to reach its peak height. The problem involves concepts from kinematics, particularly the motion of projectiles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between initial velocity, acceleration due to gravity, and time to reach peak height. Questions arise regarding the meaning of acceleration and its role in the motion of the object. There is also discussion about the components of velocity and the use of trigonometric functions in the equations of motion.

Discussion Status

The discussion is active, with participants providing insights into the equations of motion and the effects of gravity. Some participants have offered guidance on focusing on the vertical component of motion, while others are questioning the assumptions made regarding angles and components of velocity.

Contextual Notes

There is a mention of the initial vertical velocity and the need to clarify the angle of projection. The discussion also touches on the sign convention for acceleration due to gravity in the context of upward motion.

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object thrown up with initial velocity of 15 m/s and horizontal velocity of 18 m/s, how much time is required to reach the highest point of trajectory?
Vox = 18
y = yo + vosint + 1/2gt^2
 
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I assume you mean that the initial vertical component of velocity is 15 m/s.

What's the acceleration of the object? And what's the meaning of acceleration?
 
a = 9.8 ??
 
bigman8424 said:
a = 9.8 ??
The acceleration due to gravity is 9.8 m/s^2 downward. Now what's the relationship between speed, acceleration, and time?
 
bigman8424 said:
object thrown up with initial velocity of 15 m/s and horizontal velocity of 18 m/s, how much time is required to reach the highest point of trajectory?
Vox = 18
y = yo + vosint + 1/2gt^2

1st it can be assumed that y0 =0[origin]

y=V0 sin(theta) t - 1/2 gt^2... g's sign will be -ve here due to upward motion...

[i can't understand why sint is written...it should be some angle not time]

at highest point dy/dt=0 and

0= V0 sin(theta) -gt...[put t=T/2] wheret is total time of flight-----iii

V0x=V0 cos (theta)=18---------i
15cos(theta)=18

find out theta
put it in iii
T= 2V0 sin(theta)/g
 
All you need consider is the vertical component of the motion:
[tex]v_f = v_i + at[/tex]
(which is merely a restatement of what acceleration means)
 

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