To find the maximum height a ball reaches when thrown straight up, we can use the kinematic equation:

$v^2 = u^2 + 2as$

Where:

• $v$ = final velocity at maximum height (0 m/s, since the ball momentarily stops before descending)
• $u$ = initial velocity (50 m/s)
• $a$ = acceleration due to gravity (-9.8 m/s²; negative because gravity is acting in the opposite direction)
• $s$ = displacement or maximum height

Plugging in the values:

$0 = (50)^2 + 2(-9.8)s$

Simplifying:

$0 = 2500 - 19.6s$

$19.6s = 2500$

$s = \frac{2500}{19.6} \approx 127.55 \text{ meters}$

Thus, the ball will rise to a maximum height of approximately 127.55 meters.

Would you like further details or have any questions?

Related Questions:

1. How long will the ball take to reach its maximum height?
2. What is the total time the ball will be in the air before hitting the ground?
3. What will be the ball’s speed just before hitting the ground?
4. How does the maximum height change if the initial velocity is doubled?
5. How would the maximum height be affected if the gravitational acceleration is different?

Tip: When dealing with projectile motion, always identify which kinematic equation to use based on the known and unknown variables.