Projectile: time max height reached

[Newlander]

Homework Statement

"A boy throws a rock with an initial velocity of 3.13 m/s at 30.0 degrees above the horizontal. How long does it take for the rock to reach the maximum height of its trajectory?"

v_i = 3.13 m/s
t_{max height} = ?
v_f = 0 m/s (at max height)
a = -9.8 m/s²

Homework Equations

v_f = v_i + at

(Not sure!)

The Attempt at a Solution

v_f = v_i + at
t = [(v_f - v_i) / a] = [(0 m/s - 3.13 m/s)/-9.8 m/s²]
t = 0.319 s

Concern:
The correct answer is listed as 0.313 s; because my answer is so close, I wondered if this were a typo. If not, I clearly am taking the wrong approach and would appreciate some guidance.

 

[ehild]

The ball has been thrown at an angle, and the magnitude of the initial velocity is 3.13 m/s. With what initial velocity does the ball move upward?

ehild

 

[Newlander]

Hmmm . . . so, I should determine v_iy? If so, 3.13 m/s sin(30) = 1.57 m/s . . . sorry--not clear on how this will help me arrive at what's noted as the correct answer. I did plug this figure into v_f² - v_i² = 2ad, and then determined d from that . . . and then plugged that d into d = v_it + 1/2at² to determine t . . . but I ended up with the same answer, 0.319 s.

 

[ehild]

You have the formula v_f=v_i+at already. Apply it to the vertical velocity components. At the maximum height, v_y=0 (not the velocity v, as the ball keeps it horizontal velocity component during the whole flight)

ehild

 

[AJKing]

Lol, this is going to seem so simple after . . .

You used: -9.8m/s²

They used: -10m/s²

You got the correct answer under a more precise measurement.