Question

An object is going up at a velocity of 12ms-1 at a height of 32m above the ground. How long does it take for the object to reach the ground?

Answer 1

To determine the time it takes for the object to reach the ground, we can use the equations of motion under constant acceleration. In this case, the object is moving vertically against the force of gravity, so we can use the following equation:

$s=ut+\frac{1}{2}a{t}^2$

where:

• $s$ is the displacement (height above the ground), which is -32m (negative because the object is moving upward).
• $u$ is the initial velocity, which is +12 m/s (positive because the object is moving upward).
• $a$ is the acceleration due to gravity, which is approximately -9.8 m/s² (negative because it acts downward).
• $t$ is the time we need to find.

Plugging in the values, we have:

$-32=12t+\frac{1}{2}(-9.8)t^2$

Simplifying the equation, we get:

$-32=12t-4.9t^2$

Rearranging the equation to bring it to standard quadratic form:

$4.9t^2-12t-32=0$

Now, we can solve this quadratic equation to find the values of $t$. Using the quadratic formula:

t=−b±b2−4ac2a</m