# Projectile: time max height reached

## 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?"

vi = 3.13 m/s
tmax height = ?
vf = 0 m/s (at max height)
a = -9.8 m/s2

vf = vi + at

(Not sure!)

## The Attempt at a Solution

vf = vi + at
t = [(vf - vi) / 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.

## The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
ehild
Homework Helper
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

Hmmm . . . so, I should determine viy? 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 vf2 - vi2 = 2ad, and then determined d from that . . . and then plugged that d into d = vit + 1/2at2 to determine t . . . but I ended up with the same answer, 0.319 s.

ehild
Homework Helper
You have the formula vf=vi+at already. Apply it to the vertical velocity components. At the maximum height, vy=0 (not the velocity v, as the ball keeps it horizontal velocity component during the whole flight)

ehild

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.

Last edited: