To solve this problem, we can use the equations of motion for an object in free fall. Let's break down the problem into three parts:
(a) Time to reach the ground:
We can use the equation h = (1/2)gt^2, where h is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Given that the initial height is 70 m, we have 70 = (1/2)(9.8)t^2.
Simplifying the equation, we get t^2 = 70 / (0.5 * 9.8).
Solving for t, we find t = √(70 / (0.5 * 9.8)).
Evaluating the expression, we find t ≈ 3.01 seconds.
Therefore, it takes approximately 3.01 seconds for the ball to reach the ground.
(b) Initial speed of the ball:
Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s.
We can use the equation v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
Given that the final velocity is 0 m/s (at the moment the ball reaches the ground), the acceleration due to gravity is -9.8 m/s^2, and the time is 3.01 seconds, we have 0 = u + (-9.8)(3.01).
Simplifying the equation, we get u = 9.8 * 3.01.
Evaluating the expression, we find u ≈ 29.50 m/s.
Therefore, the initial speed of the ball is approximately 29.50 m/s.
(c) Velocity by which the ball will strike the ground:
Since the ball was thrown horizontally, its horizontal velocity remains constant.
The horizontal distance traveled by the ball is 45 m, and the time taken is 3.01 seconds.
Using the equation v = d / t, where v is the velocity, d is the distance, and t is the time, we have v = 45 / 3.01.
Evaluating the expression, we find v ≈ 14.95 m/s.
Therefore, the velocity by which the ball will strike the ground is approximately 14.95 m/s.
