Question

A rock is thrown straight up at 16.9 m/s.

How high does it go?
Ay = [?] m
Remember: falling = -v
lost height = -Ay
g= -9.80 m/s²

Answer 1

Using the kinematic formula for uniformly accelerated motion, the rock thrown straight up at 16.9 m/s will reach a maximum height of approximately 14.6 meters before it starts descending.

Step-by-step explanation:

The student has asked how high a rock goes when thrown straight up at 16.9 m/s. This is a Physics problem involving kinematic equations under the influence of earth's gravity. We can disregard air resistance as per the problem's conditions.

To find the maximum height (Ay) reached by the rock, we can use the formula v2 = u2 + 2as, where v is the final velocity (0 m/s at the highest point), u is the initial velocity (16.9 m/s), a is the acceleration due to gravity (-9.80 m/s2), and s is the displacement (Ay - the height reached).

Plugging the values into the formula:

0 = (16.9 m/s)2 + 2(-9.80 m/s2)Ay

So, Ay = (16.9 m/s)2 / (2 * 9.80 m/s2)

Ay = 14.6 m

Therefore, the rock will reach a maximum height of approximately 14.6 meters.