Question

A ball is launched vertically upward from ground level with an initial velocity of 30 m/s. What maximum altitude does it reach above the ground? Use g = 10 m/s².

A. 45 meters
B. 90 meters
C. 225 meters
D. 300 meters

Answer 1

Using the kinematic equation without the time variable, the ball launched vertically upward with an initial velocity of 30 m/s and under gravity of 10 m/s² reaches a maximum altitude of 45 meters above the ground.

Step-by-step explanation:

To find the maximum altitude reached by a ball launched vertically upward with an initial velocity, we can use the kinematic equation for uniformly accelerated motion, which does not require the time variable:

vf2 = vi2 + 2ad

Where:

• vf = final velocity (m/s)
• vi = initial velocity (m/s)
• a = acceleration (m/s2)
• d = displacement (m), which in this case is the altitude

Since the ball comes to a stop at the maximum altitude, the final velocity vf will be 0 m/s. Assuming the acceleration due to gravity g is -10 m/s2 (negative because it's in the opposite direction to the initial velocity), and the initial velocity vi is 30 m/s, we get:

0 = (30 m/s)2 + 2(-10 m/s2)(d)

Solving for d gives us:

d = (30 m/s)2 / (2 * 10 m/s2)

d = 900 m2/s2 / 20 m/s2

d = 45 m

Therefore, the maximum altitude the ball reaches is 45 meters, which corresponds to option A.