A boy throws a rock with an initial velocity of 2.15 m/s at 30.0° above the horizontal. If air resistance is negligible, how long does it take for the rock to reach the maximum height of its trajectory?

t = 0.11 second

The time taken to reach the maximum height of the trajectory is 0.11 second

Explanation:

Taking the vertical component of the initial velocity;

Vy = Vsin∅

Initial velocity V = 2.15 m/s

Angle ∅ = 30°

Vy = 2.15sin30 = 2.15 * 0.5

Vy = 1.075 m/s

The height of the rock at time t during the flight is;

From the equation of motion;

h (t) = Vy*t - 0.5gt^2

g = acceleration due to gravity = 9.8m/s^2

Substituting the given values;

h (t) = 1.075t - 0.5 (9.8) t^2

h (t) = 1.075t - 4.9t^2

The rock is at maximum height when dh/dt = 0;

dh (t) / dt = 1.075 - 9.8t = 0

1.075 - 9.8t = 0

9.8t = 1.075

t = 1.075/9.8

t = 0.109693877551 s

t = 0.11 second

The time taken to reach the maximum height of the trajectory is 0.11 second