An object is thrown vertically from a height of #6 m# at # 3 m/s#. How long will it take for the object to hit the ground?

Question

An object is thrown vertically from a height of #6   m# at #  3  m/s#.  How long will it take for the object to hit the ground?

Layla Ahmed · Accepted Answer

Use, #s=ut -1/2 g t^2# to solve it. Total displacement in reaching ground will be #-6m# So,if it takes time #t# then, #-6=3t - 1/2×9.8×t^2# #9.8t^2-6t-12=0# So,#t=1.44s#

Benjamin Asiedu · Answer

undefined To calculate the time it takes for the object to hit the ground, we can use the following kinematic equation for vertical motion:

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

Where:
- $ s $ is the displacement (height) of the object, which is -6 m (negative because it's downward).
- $ u $ is the initial velocity, which is 3 m/s (upward).
- $ g $ is the acceleration due to gravity, which is approximately $ 9.8 \, \text{m/s}^2 $ (downward).
- $ t $ is the time taken.

Plugging in the values:

$$ -6 = (3)t + \frac{1}{2}(9.8)t^2 $$

Rearranging the equation and solving for $ t $:

$$ 4.9t^2 + 3t - 6 = 0 $$

Using the quadratic formula:

$$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Where:
- $ a = 4.9 $
- $ b = 3 $
- $ c = -6 $

$$ t = \frac{-3 \pm \sqrt{(3)^2 - 4(4.9)(-6)}}{2(4.9)} $$

$$ t \approx \frac{-3 \pm \sqrt{9 + 117.6}}{9.8} $$

$$ t \approx \frac{-3 \pm \sqrt{126.6}}{9.8} $$

$$ t \approx \frac{-3 \pm 11.26}{9.8} $$

$$ t_1 \approx \frac{-3 + 11.26}{9.8} $$
$$ t_2 \approx \frac{-3 - 11.26}{9.8} $$

$$ t_1 \approx \frac{8.26}{9.8} \approx 0.843 \text{ seconds} $$
$$ t_2 \approx \frac{-14.26}{9.8} \approx -1.454 \text{ seconds} $$

Since time cannot be negative, we take the positive value:

$$ t \approx 0.843 \text{ seconds} $$

So, it will take approximately 0.843 seconds for the object to hit the ground.