Final answer:
The ball thrown straight up takes 2 seconds to reach the highest point, where its velocity is 0 m/s. By using the kinematic equation for constant acceleration, we calculate that the ball reaches a height of around 78.4 meters, with the nearest given option being 80 meters, hence the answer is C) 80 meters.
Step-by-step explanation:
To determine how high a ball thrown straight up went if it took 4 seconds to return to its launching point, we can utilize kinematic equations that describe projectile motion under constant acceleration, in this case, due to gravity (approximately 9.8 m/s2 on Earth). The total time of flight is the sum of the time the ball takes to reach its highest point (apex) and the time it takes to come back down. Since the motion is symmetrical, it will take half of the total time to reach the apex, which means 2 seconds (half of 4 seconds).
Using the kinematic equation h = (vi × t) - (0.5 × g × t2), where h is the height, vi is the initial velocity, g is the acceleration due to gravity (9.8 m/s2), and t is the time, we can find the height at the apex. As the ball reaches its apex, its velocity becomes 0 m/s. Since the ball takes 2 seconds to reach the apex, we use this time in our equation with g being 9.8 m/s2.
h = 0 - (0.5 × 9.8 m/s2 × 22)
= - (0.5 × 9.8 m/s2 × 4)
= - (19.6 m/s2)×4
= - 78.4 m. However, since height can't be negative, we take the absolute value and get 78.4 meters as the answer, which means the correct choice is option C) 80 meters (rounded to the nearest tenth).