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Highest point of trajectory

  • Thread starter bigman8424
  • Start date
25
0
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
 

Doc Al

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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?
 
25
0
a = 9.8 ??
 

Doc Al

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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?
 
29
0
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 cant 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
 

Doc Al

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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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