Question

what is the velocity of a ball that has been dropped off a cliff after 1, 2, 3, 4, and 5 seconds? (assume g is rounded to -10 m/s²)? -20, -45, -80, -125 m -10,-10,-10,-10,-10 m/s -10, -20, -30, -40, -50 m/s -5, -10, -15, -20, -25 m/s

Explanation:

Step1: Recall the velocity - time formula

The formula for the velocity of a freely - falling object is $v = v_0+gt$, where $v_0$ is the initial velocity, $g$ is the acceleration due to gravity, and $t$ is the time. Since the ball is dropped, $v_0 = 0$. Given $g=- 10\ m/s^2$.

Step2: Calculate velocity at $t = 1\ s$

Substitute $t = 1\ s$, $v_0 = 0$ and $g=-10\ m/s^2$ into the formula $v=v_0 + gt$. We get $v=0+( - 10)\times1=-10\ m/s$.

Step3: Calculate velocity at $t = 2\ s$

Substitute $t = 2\ s$, $v_0 = 0$ and $g=-10\ m/s^2$ into the formula. $v=0+( - 10)\times2=-20\ m/s$.

Step4: Calculate velocity at $t = 3\ s$

Substitute $t = 3\ s$, $v_0 = 0$ and $g=-10\ m/s^2$ into the formula. $v=0+( - 10)\times3=-30\ m/s$.

Step5: Calculate velocity at $t = 4\ s$

Substitute $t = 4\ s$, $v_0 = 0$ and $g=-10\ m/s^2$ into the formula. $v=0+( - 10)\times4=-40\ m/s$.

Step6: Calculate velocity at $t = 5\ s$

Substitute $t = 5\ s$, $v_0 = 0$ and $g=-10\ m/s^2$ into the formula. $v=0+( - 10)\times5=-50\ m/s$.

Answer:

The velocities at 1, 2, 3, 4, and 5 seconds are - 10, - 20, - 30, - 40, - 50 m/s respectively, so the correct option is the one with values - 10, - 20, - 30, - 40, - 50 m/s.