A boy throws a rock with an initial velocity of 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?

Answer: 0.5m/s

This value may vary depending on the initial velocity value.

Explanation:

The question is incomplete due to absence of the initial velocity (U) of the rock. Taking the initial velocity as any value say 10m/s at 30° above the horizontal.

Time it will take to reach maximum height (Tmax) will be Usin (theta) / g

Where U is the initial velocity = 10m/s theta = 30° g is the acceleration due to gravity = 10m/s²

Substituting the values in the formula we have;

Tmax = 10sin30°/10

Tmax = sin 30°

Tmax = 0.5second

Therefore, it will take for the rock 0.5s to reach the maximum height of its trajectory if the velocity is 10m/s.

The time varies though depending on the value of the initial velocity.