Question:

An object is thrown straight up into the air with an initial velocity of 30 m/s. The acceleration due to gravity is -9.8 m/s^2. Calculate the maximum height reached by the object.

Answer:

Given:

Initial velocity, u = 30 m/s (upward)

Acceleration due to gravity, a = -9.8 m/s^2 (downward)

We are asked to find the maximum height reached by the object.

Step 1: Identify the knowns and unknowns:

• Initial velocity (u) = 30 m/s (known)
• Final velocity (v) = 0 m/s (at maximum height, the object momentarily comes to rest)
• Acceleration due to gravity (a) = -9.8 m/s^2 (known)
• Displacement (s) = ? (unknown)

Step 2: Use the equation of motion relating final velocity, initial velocity, acceleration, and displacement:

v² = u² + 2as

Substituting the known values:

0 = (30)² + 2(-9.8)s

Step 3: Solve for displacement (s):

900 = -19.6s

Dividing both sides of the equation by -19.6:

s = 900 / -19.6

s ≈ -45.92 m

The negative sign indicates that the object is moving in the opposite direction to the initial motion (downward). However, for the maximum height calculation, we need the magnitude of the displacement. Therefore, we take the absolute value of the displacement.

Step 4: Calculate the magnitude of the maximum height:

Height = |-45.92| = 45.92 m

The maximum height reached by the object is approximately 45.92 meters.