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

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

are there options? cause you ain’t. real specific