Answer:
27 m/s
Explanation:
Positive direction: up
v₀ = 16 m/s, y₀ = 24 m, g = -9.8 m/s², y = 0 (ground)
Use v² - v₀² = 2g(y - y₀)
v² - 16² = 2*(-9.8)*(0 - 24)
v² = 256 + 470.4 = 726.4
so v = 27 m/s
Answer:
About 26.95 m/s.
Explanation:
We can let downwards be the positive direction and upwards be the negative direction.
The ball is thrown upwards at a speed of 16 m/s. (Hence, the initial velocity is expressed as -16 m/s.)
Find the time it took for the ball to travel to its highest point. At this point, its velocity is zero. Find time t using a kinematic equation. In the vertical direction, the acceleration is gravity:
[tex]\displaystyle \begin{aligned} v_y & = v_i + gt \\ \\ (0) & = (-16\text{ m/s}) + (9.8\text{ m/s$^2$})t \\ \\ t &= \frac{80}{49} \text{ s}\end{aligned}[/tex]
Find the distance it traveled during this time by using the following kinematic equation:
[tex]\displaystyle \begin{aligned} y & = v_{i}t + \frac{1}{2}gt^2 \\ \\ & = (-16\text{ m/s})\left(\frac{80}{49}\text{ s}\right) + \frac{1}{2}(9.8\text{ m/s$^2$})\left(\frac{80}{49}\text{ s}\right)^2 \\ \\ & \approx -13.0612 \text{ m}\end{aligned}[/tex]
Hence, the ball traveled a total of 13.0612 meters when thrown into the air.
Thus, its total height is:
[tex]\displaystyle h_\text{max} = 24 \text{ m} + 13.0612 \text{ m} = 37.0612\text{ m}[/tex]
At this point, it will fall back down with an initial velocity of zero.
Find the time it took for the ball to hit the ground below the cliff:
[tex]\displaystyle \begin{aligned} y & = v_{i}t + \frac{1}{2}gt^2 \\ \\ (37.0612\text{ m}) & = (0\text{ m/s})t + \frac{1}{2}(9.8\text{ m/s$^2$})t^2 \\ \\ t & \approx 2.75\text{ s} \end{aligned}[/tex]
Find its final velocity using a kinematic equation:
[tex]\displaystyle \begin{aligned} v_{f} & = v_{i} + gt \\ \\ & = (0\text{ m/s}) + (9.8\text{ m/s$^2$})(2.75\text{ s}) \\ \\ & = 26.95\text{ m/s}\end{aligned}[/tex]
In conclusion, the final velocity of the ball is about 26.95 m/s.
