Question

a ball is thrown horizontally off a roof as shown below: 5. how long does it take to reach the ground? 6. what is v_y when it hits the ground? 7. what is v_x when it hits the ground? 8. what horizontal distance has it traveled by the time it lands? 9. what is the balls acceleration as it falls?

Explanation:

Step1: Find time to reach ground

Use vertical - motion equation $h = v_{0y}t+\frac{1}{2}gt^{2}$. Since $v_{0y} = 0$ (thrown horizontally), $h=\frac{1}{2}gt^{2}$. Given $h = 28m$ and $g = 9.8m/s^{2}$, we can solve for $t$.
$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times28}{9.8}}\approx2.39s$

Step2: Find vertical velocity $v_y$ when it hits the ground

Use the equation $v_y=v_{0y}+gt$. Since $v_{0y} = 0$, $v_y = gt$. Substituting $g = 9.8m/s^{2}$ and $t\approx2.39s$, we get $v_y=9.8\times2.39 = 23.42m/s$

Step3: Find horizontal velocity $v_x$ when it hits the ground

In horizontal direction, there is no acceleration ($a_x = 0$), so $v_x$ remains constant. Given $v_x=9m/s$ initially, $v_x = 9m/s$ when it hits the ground.

Step4: Find horizontal distance $x$

Use the equation $x = v_x t$. Substituting $v_x = 9m/s$ and $t\approx2.39s$, we get $x=9\times2.39 = 21.51m$

Step5: Find acceleration as it falls

The acceleration of the ball as it falls is the acceleration due to gravity, $a = g=9.8m/s^{2}$

Answer:

1. $t\approx2.39s$
2. $v_y = 23.42m/s$
3. $v_x = 9m/s$
4. $x = 21.51m$
5. $a = 9.8m/s^{2}$