Question

A tennis ball is thrown straight up over a cliff with a speed of 39.706 m/s. If the cliff is 20.776m high, what is the time it takes the ball to hit the ground?

Answer 1

Answer 2

To find the time it takes for the tennis ball to hit the ground, we can use the kinematic equation for vertical motion:

h = vi*t + (1/2) * g * t^2

where: h is the initial height (positive when above the ground), vi is the initial vertical velocity (positive when upward), g is the acceleration due to gravity (approximately -9.81 m/s^2, negative because it acts downward), t is the time.

At the highest point, the vertical velocity becomes zero before the ball starts descending, so vi = 0 m/s.

Given: vi = 39.706 m/s (upward) h = 20.776 m g = -9.81 m/s^2

Let's solve for the time (t):

0 = 39.706 * t + (1/2) * (-9.81) * t^2

Simplify the equation:

-4.905 * t^2 + 39.706 * t = 0

Now, we can solve for t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

where a = -4.905, b = 39.706, and c = 0.

t = (-(39.706) ± √((39.706)^2 - 4 * (-4.905) * 0)) / 2 * (-4.905)

t = (-39.706 ± √(1574.372 - 0)) / (-9.81)

t = (-39.706 ± √1574.372) / (-9.81)

Now, calculate the two possible values of t:

t₁ = (-39.706 + √1574.372) / (-9.81) t₁ = (-39.706 + 39.703) / (-9.81) t₁ ≈ -0.003 / (-9.81) t₁ ≈ 0.000305 seconds (approximately)

t₂ = (-39.706 - √1574.372) / (-9.81) t₂ = (-39.706 - 39.703) / (-9.81) t₂ ≈ -79.409 / (-9.81) t₂ ≈ 8.096 seconds (approximately)

The negative value for t₁ doesn't make sense in this context, so we discard it. The ball hits the ground approximately 8.096 seconds after it is thrown.