Highest point of trajectory

1. Jun 17, 2005

bigman8424

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

2. Jun 17, 2005

Staff: Mentor

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. Jun 17, 2005

bigman8424

a = 9.8 ??

4. Jun 18, 2005

Staff: Mentor

The acceleration due to gravity is 9.8 m/s^2 downward. Now what's the relationship between speed, acceleration, and time?

5. Jun 18, 2005

aura

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

6. Jun 18, 2005

Staff: Mentor

All you need consider is the vertical component of the motion:
$$v_f = v_i + at$$
(which is merely a restatement of what acceleration means)