An object is thrown vertically from a height of #9 m# at #23 ms^-1#. How long will it take for the object to hit the ground?

1 Answer
Jan 14, 2016


Assuming the initial velocity is downward, the object will hit the ground after #0.36 s#.


The question is a little ambiguous: is the object initially thrown upward, downward, horizontally or at an angle? I will assume it is thrown downward, but as a bonus will calculate the time taken if it is thrown upward.

Assuming it is thrown downward, the initial velocity of #23 ms^-1# and the acceleration due to gravity of #9.8 ms^-2# are in the same direction. We can use the following formula, but it will yield a quadratic equation we need to solve:

# d = ut + 1/2 at^2#

Substituting in what we know, we get:

# 9 = 23t + 1/2 9.8t^2#

Rearrange to standard form for a quadratic equation:

# 1/2 9.8t^2 + 23t - 9 = 0#

I won't do it here, but I'd solve this with the quadratic formula, though there are other approaches. The roots are #t = 0.36 s and -5.06 s#. We'll ignore the negative root and state that it takes #0.36 s#. for the stone to reach the ground.

(as a quick check for sensibleness you could calculate that a stone simply dropped from 9 m would take #1.35 s# to hit the ground: since in this case it was thrown downward quite fast (#23 ms^-1 =# about #83 kmh^-h#) the smaller answer makes sense.)

(If we assume instead that the stone was thrown upward, we simply give the initial velocity a negative sign, since it is in the opposite direction to the acceleration. Working it through gives #5.06 s#. Hmm... that number seems oddly familiar!)