Answer:
Time = 5.1 seconds
speed = 10 m/s
Step-by-step explanation:
Let's use the kinematic equations to solve the problem.
At the maximum height, the ball has zero velocity, so we can use the following kinematic equation:
v_f^2 = v_i^2 + 2ad
where v_f is the final velocity (which is zero), v_i is the initial velocity, a is the acceleration due to gravity (-10 m/s^2), and d is the displacement (which is 50 m).
Solving for v_i, we get:
v_i = sqrt(2ad) = sqrt(2 x (-10 m/s^2) x 50 m) = 10 m/s
Therefore, the speed of the ball when it was thrown vertically upward was 10 m/s.
To calculate the time taken to reach the maximum height, we can use the following kinematic equation:
d = v_i t + 1/2at^2
where d is the displacement (which is 50 m), v_i is the initial velocity (which is 10 m/s), a is the acceleration due to gravity (-10 m/s^2), and t is the time taken.
Solving for t, we get:
t = (-v_i ± sqrt(v_i^2 - 4ad)) / 2a
Since the ball was thrown vertically upward, we use the positive root:
t = (-10 m/s + sqrt((10 m/s)^2 - 4 x (-10 m/s^2) x 50 m)) / 2 x (-10 m/s^2) ≈ 5.1 s
Therefore, the time taken for the ball to reach the maximum height is approximately 5.1 seconds.