How Long to Reach Peak Height for a Thrown Object?

In summary, the object thrown with an initial vertical velocity of 15 m/s and horizontal velocity of 18 m/s will reach its highest point in a time of 1.53 seconds. This can be calculated by finding the angle of the initial velocity (theta) using the horizontal velocity and then plugging it into the equation T = 2V0 sin(theta)/g. The acceleration due to gravity is 9.8 m/s^2.
  • #1
bigman8424
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
 
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  • #2
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?
 
  • #3
a = 9.8 ??
 
  • #4
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?
 
  • #5
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
 
  • #6
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)
 

FAQ: How Long to Reach Peak Height for a Thrown Object?

What is the highest point of trajectory?

The highest point of trajectory is the point in the trajectory of a projectile where it has reached its maximum height before descending back to the ground.

How is the highest point of trajectory calculated?

The highest point of trajectory is calculated using the equation h = (v02sin2θ)/2g, where h is the maximum height, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

What factors affect the highest point of trajectory?

The highest point of trajectory is affected by the initial velocity, launch angle, air resistance, and the acceleration due to gravity.

Can the highest point of trajectory be higher than the initial height?

Yes, the highest point of trajectory can be higher than the initial height if the projectile is launched at a high enough angle and with enough initial velocity.

How is the highest point of trajectory used in real-world applications?

The knowledge of the highest point of trajectory is used in fields such as ballistics, sports, and engineering to calculate the trajectory of projectiles and ensure accurate and precise outcomes.

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