A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45 degrees above the horizontal. With what speed was it thrown?Neglect air friction and express your answer in m/s.

To solve this problem, we need to use the equations of motion.

Step 1: Calculate the time it takes for the ball to hit the ground.

We know that the height of the hill is 20m and the acceleration due to gravity is approximately 9.8 m/s². We can use the equation of motion h = 0.5gt² to find the time it takes for the ball to hit the ground.

20 = 0.5 * 9.8 * t² Solving for t, we get t = sqrt(20 / 4.9) = 2.02 seconds.

Step 2: Calculate the horizontal distance the ball travels.

We know that the ball hits the ground at an angle of 45 degrees. This means that the horizontal and vertical components of the ball's velocity are equal when it hits the ground.

We can use the equation v = gt to find the vertical component of the ball's velocity when it hits the ground.

v = 9.8 * 2.02 = 19.8 m/s.

Since the horizontal and vertical components of the ball's velocity are equal, the horizontal component of the ball's velocity is also 19.8 m/s.

Step 3: Calculate the initial speed of the ball.

We know that the ball was thrown horizontally, so the initial vertical component of the ball's velocity is 0. The horizontal component of the ball's velocity is the same throughout its flight, so the initial speed of the ball is the same as the horizontal component of the ball's velocity when it hits the ground.

Therefore, the ball was thrown with a speed of 19.8 m/s.