A ball is thrown horizontally of a cliff. If the initial speed of the ball is 17 m/s and the cliff is 77 m high, how far from the base of the cliff will the ball land in the water below? Calculate the answer in meters (m) and rounded to three significant figures.

S=at/2.

77=5t^2

t=15.4s

range=vt

R=261.8m

To find the distance from the base of the cliff where the ball will land, we need to determine the time it takes for the ball to fall from the height of the cliff.

First, let's calculate the time it takes for the ball to hit the ground using the equation:

\[
h = \frac{1}{2} g t^2
\]

where h is the height of the cliff (77 m), g is the acceleration due to gravity (9.8 m/s²), and t is the time. Rearranging the equation to solve for t, we have:

\[
t = \sqrt{\frac{2h}{g}}
\]

Plugging in the values, we get:

\[
t = \sqrt{\frac{2 \cdot 77 \, \text{m}}{9.8 \, \text{m/s}^2}}
\]

Simplifying the equation:

\[
t = \sqrt{\frac{154 \, \text{m}}{9.8 \, \text{m/s}^2}}
\]

\[
t \approx \sqrt{15.714}
\]

\[
t \approx 3.965 \, \text{s}
\]

Now that we know the time it takes for the ball to hit the ground, we can calculate the horizontal distance it covers using the equation:

\[
\text{distance} = \text{velocity} \times \text{time}
\]

In this case, the velocity is given as the initial speed of the ball, which is 17 m/s, and the time is 3.965 s. Plugging in the values, we get:

\[
\text{distance} = 17 \, \text{m/s} \times 3.965 \, \text{s}
\]

\[
\text{distance} \approx 67.405 \, \text{m}
\]

Therefore, the ball will land approximately 67.405 meters from the base of the cliff.