S=at/2.
77=5t^2
t=15.4s
range=vt
R=261.8m
77=5t^2
t=15.4s
range=vt
R=261.8m
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/sĀ²), 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.