Question

I throw a ball directly upward at t=0 at an initial speed of 14.3 m/s. What is the maximum height the ball reaches above where it leaves my hand? Ignore air resistance and take g = 9.8 m/s^2. What is the max height? What time does it pass through earlier time half of the maximum height? Later time at half of max height?

a. The maximum height is given by H_max = 0.5 * g * t_max^2, with t_earlier = t_max/2 and t_later = 2 * t_max.
b. The maximum height is given by H_max = 0.5 * g * t_max^2, with t_earlier = 2 * t_max and t_later = t_max/2.
c. The maximum height is given by H_max = g * t_max, with t_earlier = t_max/2 and t_later = 2 * t_max.
d. The maximum height is given by H_max = g * t_max, with t_earlier = 2 * t_max and t_later = t_max/2.

Answer 1

The maximum height the ball reaches is 10.258 meters. The earlier time at half of the maximum height is 0.71 seconds and the later time is 2.85 seconds.

Step-by-step explanation:

The maximum height the ball reaches can be found using the equation H_max = (V_0^2) / (2g), where H_max is the maximum height, V_0 is the initial velocity, and g is the acceleration due to gravity. Plugging in the given values, we have H_max = (14.3^2) / (2 * 9.8) = 10.258 meters.

The time it takes for the ball to pass through earlier time half of the maximum height is given by t_earlier = sqrt(2 * H_max / g) / 2 = sqrt(2 * 10.258 / 9.8) / 2 = 0.71 seconds.

The time it takes for the ball to pass through later time half of the maximum height is given by t_later = 2 * sqrt(2 * H_max / g) = 2 * sqrt(2 * 10.258 / 9.8) = 2.85 seconds.