Question

A ball is thrown horizontally from the top of a cliff with an initial speed of 10 m/s. The cliff has a height of 20 meters.

a) What is the time it takes for the ball to hit the ground? b) What is the horizontal displacement of the ball when it hits the ground?

Use a constant acceleration due to gravity of g = 9.8 m/s².

Answer

a) To find the time it takes for the ball to hit the ground, we can use the equation for displacement in the vertical direction:

Δy = v₀y · t + 0.5 · a · t²

Since the ball is thrown horizontally, the initial vertical velocity (v₀y) is 0 m/s, and the acceleration in the vertical direction (a) is equal to the acceleration due to gravity (g = 9.8 m/s²). The initial vertical displacement is 20 m, and we want to find the time (t).

Plugging in the known values into the equation:

-20 m = 0 m/s · t + 0.5 · (9.8 m/s²) · t²

Simplifying the equation:

-20 m = 4.9 m/s² · t²

Rearranging the equation:

t² = -20 m / (4.9 m/s²)

t² = 4.08163 s²

Taking the square root of both sides:

t = 2.0208 s

Therefore, the time it takes for the ball to hit the ground is approximately 2.02 seconds.

b) To find the horizontal displacement of the ball when it hits the ground, we can use the equation for horizontal motion:

Δx = v₀x · t

Since the ball is thrown horizontally, the initial horizontal velocity (v₀x) is 10 m/s, and we already know the time (t) from part (a).

Plugging in the known values into the equation:

Δx = (10 m/s) · (2.0208 s)

Δx ≈ 20.21 m

Therefore, the horizontal displacement of the ball when it hits the ground is approximately 20.21 meters.