An object is thrown vertically from a height of #2 m# at #5m/s#. How long will it take for the object to hit the ground?

Question

An object is thrown vertically from a height of #2 m# at #5m/s#.  How long will it take for the object to hit the ground?

Nathan Ashcroft · Accepted Answer

It will take 1.32 seconds for the object to hit the ground.

If a body is projected or thrown vertically up from a height 'h' with a velocity 'u' then time taken by the body to reach the ground is #t = [u+sqrt(u^2 + 2gh)]//g# Consider 'g' acceleration due to gravity as #9.8m/s^2# By substituting the values in above equation we get #=> t= [5+sqrt((5)^2 + 2xx9.8xx2)]//9.8# #:.t=1.32 seconds#

Adam Arnold · Answer

undefined To find the time it takes for the object to hit the ground, you can use the equation:

$$ t = \sqrt{\frac{2h}{g}} $$

where $ h $ is the initial height (2 m) and $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $).

Substituting the given values:

$$ t = \sqrt{\frac{2 \times 2}{9.8}} $$

$$ t \approx \sqrt{\frac{4}{9.8}} $$

$$ t \approx \sqrt{0.408} $$

$$ t \approx 0.64 \, \text{s} $$

So, it will take approximately 0.64 seconds for the object to hit the ground.

Layla Ahmed · Answer

undefined To find the time it takes for the object to hit the ground, you can use the kinematic equation for vertical motion under gravity:

$$ h = h_0 + v_0 t - \frac{1}{2}gt^2 $$

where:
$ h $ = final height (which is 0 since the object hits the ground)
$ h_0 $ = initial height (given as 2 m)
$ v_0 $ = initial velocity (given as 5 m/s)
$ g $ = acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $)
$ t $ = time

Substituting the given values into the equation and solving for $ t $, you can find the time it takes for the object to hit the ground.