A ball is thrown horizontally from the top of

Question

a building 150 m high. The ball strikes the
ground 62 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s

Answer 1

To find the speed of the ball just before it strikes the ground, we can use the equation of motion for horizontal motion:

distance = speed * time

Since the ball is thrown horizontally, the time of flight is the same as if it were dropped from the same height. We can use the equation for vertical motion:

distance = 1/2 * acceleration * time^2

In this case, the initial velocity in the vertical direction is 0, since the ball is only moving horizontally. The acceleration due to gravity is -9.8 m/s^2 because it acts downwards.

Using the equation for distance in vertical motion, we can solve for time:

150 = 1/2 * -9.8 * time^2

Simplifying the equation:

300 = -9.8 * time^2

Dividing by -9.8:

time^2 = -30.61

Since time cannot be negative, we take the positive square root:

time = √30.61 ≈ 5.52 s

Now, we can use this time to find the horizontal speed of the ball. The horizontal distance is given as 62 m. Using the equation for distance in horizontal motion:

62 = speed * time

Speed = 62 / time
= 62 / 5.52
≈ 11.23 m/s

Therefore, the speed of the ball just before it strikes the ground is approximately 11.23 m/s.